Page No 13.3:
Question 1:
Evaluate the following:
(i) i457
(ii) i528
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
Answer:
Page No 13.31:
Question 1:
Express the following complex numbers in the standard form a + i b:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
Answer:
Page No 13.31:
Question 2:
Find the real values of x and y, if
(i)
(ii)
(iii)
(iv)
Answer:
Page No 13.31:
Question 3:
Find the conjugates of the following complex numbers:
(i) 4 − 5 i
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
Page No 13.32:
Question 4:
Find the multiplicative inverse of the following complex numbers:
(i) 1 − i
(ii)
(iii) 4 − 3i
(iv)
Answer:
Page No 13.32:
Question 5:
If
Answer:
Page No 13.32:
Question 6:
If find
(i) Re
(ii) Im
Answer:
Page No 13.32:
Question 7:
Find the modulus of
Answer:
Page No 13.32:
Question 8:
If , prove that x2 + y2 = 1
Answer:
Page No 13.32:
Question 9:
Find the least positive integral value of n for which is real.
Answer:
Page No 13.32:
Question 10:
Find the real values of θ for which the complex number is purely real.
Answer:
Page No 13.32:
Question 11:
Find the smallest positive integer value of m for which is a real number.
Answer:
Page No 13.32:
Question 12:
If , find (x, y).
Answer:
Also,
It is given that,
Thus, (
x,
y) = (0, −2).
Page No 13.32:
Question 13:
If , find x + y.
Answer:
It is given that,
Thus,
x + y =
.
Page No 13.32:
Question 14:
If , find (a, b).
Answer:
It is given that,
Thus, (
a,
b) = (1, 0).
Page No 13.32:
Question 15:
If , find the value of .
Answer:
Thus,
.
Page No 13.32:
Question 16:
Evaluate the following:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
Page No 13.32:
Question 17:
For a positive integer n, find the value of .
Answer:
Thus, the value of
is 2
n.
Page No 13.33:
Question 18:
If , then show that .
Answer:
Hence,
.
Page No 13.33:
Question 19:
Solve the system of equations .
Answer:
Let .
Then ,
and
According to the question,
Thus, .
Page No 13.33:
Question 20:
If is purely imaginary number (), find the value of .
Answer:
Let .
Then,
If is purely imaginary number, then
Thus, the value of is 1.
Page No 13.33:
Question 21:
If z1 is a complex number other than −1 such that and , then show that the real parts of z2 is zero.
Answer:
Let .
Then,
Now,
Thus, the real parts of z2 is zero.
Page No 13.33:
Question 22:
If , find z.
Answer:
Let .
Then,
Thus,
Page No 13.33:
Question 23:
Solve the equation .
Answer:
Let .
Then,
âThus,
Page No 13.33:
Question 24:
What is the smallest positive integer n for which ?
Answer:
Thus, the smallest positive integer
n for which
is 2.
Page No 13.33:
Question 25:
If z1, z2, z3 are complex numbers such that , then find the value of .
Answer:
Thus, the value of
is 1.
Page No 13.33:
Question 26:
Find the number of solutions of
Answer:
Let .
Then,
For
âThus, there are infinitely many solutions of the form .
Page No 13.39:
Question 1:
Find the square root of the following complex numbers:
(i) −5 + 12i
(ii) −7 − 24i
(iii) 1 − i
(iv) −8 − 6i
(v) 8 −15i
(vi)
(vii)
(viii) 4i
(ix) −i
Answer:
Page No 13.4:
Question 2:
Show that 1 + i10 + i20 + i30 is a real number.
Answer:
Page No 13.4:
Question 3:
Find the values of the following expressions:
(i) i49 + i68 + i89 + i110
(ii) i30 + i80 + i120
(iii) i + i2 + i3 + i4
(iv) i5 + i10 + i15
(v)
(vi) 1+ i2 + i4 + i6 + i8 + ... + i20
(vii) (1 + i)6 + (1 − i)3
Answer:
(vii) (1 +
i)
6 + (1 −
i)
3
= [(1 +
i)
2]
3 + (1 −
i)
3
= [1
2 +
i2 + 2
i]
3 + (1
3 −
i3 + 3
i2 − 3
i)
= [1 − 1
+ 2
i]
3 + (1 +
i − 3
− 3
i) [âµ
i2 = −1,
i3 = −
i]
= (2
i)
3 + (−2
− 2
i)
= 8
i3 − 2
− 2
i
= −8
i − 2
− 2
i [âµ
i3 = −
i]
= −10
i − 2
Page No 13.57:
Question 1:
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1 + i
(ii)
(iii) 1 − i
(iv)
(v)
(vi)
(vii)
(viii)
Answer:
Page No 13.57:
Question 2:
Write (i25)3 in polar form.
Answer:
Let
.
Then,
.
Let θ be the argument of
z and α be the acute angle given by
.
Then,
Clearly,
z lies in fourth quadrant. So, arg(
z) =
.
∴ the polar form of
z is
.
Thus, the polar form of (
i25)
3 is
.
Page No 13.57:
Question 3:
Express the following complex in the form r(cos θ + i sin θ):
(i) 1 + i tan α
(ii) tan α − i
(iii) 1 − sin α + i cos α
(iv)
Answer:
Page No 13.57:
Question 4:
If z1 and z2 are two complex numbers such that and arg(z1) + arg(z2) = , then show that .
Answer:
Let θ1 be the arg(z1) and θ2 be the arg(z2).
It is given that and arg(z1) + arg(z2) = .
Since, z1 is a complex number.
Hence, .
Page No 13.57:
Question 5:
If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, prove that .
Answer:
Given that z1, z2 and z3, z4 are two pairs of conjugate complex numbers.
Then,
and
Hence, .
Page No 13.58:
Question 6:
Express in polar form.
Answer:
Page No 13.62:
Question 1:
The value of is
(a) 2
(b) 0
(c) 1
(d) i
Answer:
(b) 0
(1+ i) (1 + i2) (1 + i3) (1 + i4)
= (1+ i) (1 1) (1 i) (1 + 1) (i2 = 1, i3 = i and i4 = 1)
= (1 + i) (0) (1 i) (2)
= 0
Page No 13.62:
Question 2:
If is a real number and 0 < θ < 2π, then θ =
(a) π
(b)
(c)
(d)
Answer:
(a) π
Given:
is a real number
On rationalising, we get,
For the above term to be real, the imaginary part has to be zero.
For this to be zero,
sin = 0
= 0,
But
Hence,
Page No 13.62:
Question 3:
If is equal to
(a)
(b)
(c)
(d)
(e)
Answer:
(c) a2+ b2
(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib
Taking modulus on both the sides, we get:
Squaring on both the sides, we get:
2
Page No 13.62:
Question 4:
If then possible value of is
(a)
(b)
(c) x + iy
(d) x − iy
(e)
Answer:
(d) x iy
Page No 13.62:
Question 5:
If , then
(a)
(b)
(c)
(d)
Answer:
(d)
Page No 13.62:
Question 6:
The polar form of (i25)3 is
(a)
(b) cos π + i sin π
(c) cos π − i sin π
(d)
Answer:
(d)
(i25)3 = (i)75
= (i)418+ 3
= (i)3
= i (i4 = 1)
Modulus, r =
Polar form = r (cos + i sin )
= cos+i sin
= cosi sin
Page No 13.62:
Question 7:
If i2 = −1, then the sum i + i2 + i3 +... upto 1000 terms is equal to
(a) 1
(b) −1
(c) i
(d) 0
Answer:
(d) 0
Page No 13.63:
Question 8:
If , then the value of arg (z) is
(a) π
(b)
(c)
(d)
Answer:
(c)
z =
Rationalising z, we get,
Page No 13.63:
Question 9:
If a = cos θ + i sin θ, then
(a)
(b) cot θ
(c)
(d)
Answer:
(c)
Page No 13.63:
Question 10:
If (1 + i) (1 + 2i) (1 + 3i) .... (1 + ni) = a + ib, then 2.5.10.17.......(1+n2)=
(a) a − ib
(b) a2 − b2
(c) a2 + b2
(d) none of these
Answer:
(c) a2+ b2
(1 + i)(1 + 2i)(1 + 3i) ......(1 + ni) = a + ib
Taking modulus on both the sides, we get,
Squaring on both the sides, we get:
2×5×10×.....(1 + n2) = a2 + b2
Page No 13.63:
Question 11:
If is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(a)
Taking modulus on both the sides, we get:
Page No 13.63:
Question 12:
The principal value of the amplitude of (1 + i) is
(a)
(b)
(c)
(d) π
Answer:
(a)
Let z = (1+i)
Therefore, arg (z) =
Page No 13.63:
Question 13:
The least positive integer n such that is a positive integer, is
(a) 16
(b) 8
(c) 4
(d) 2
Answer:
Page No 13.63:
Question 14:
If z is a non-zero complex number, then is equal to
(a)
(b)
(c)
(d) none of these
Answer:
(a)
Page No 13.63:
Question 15:
If a = 1 + i, then a2 equals
(a) 1 − i
(b) 2i
(c) (1 + i) (1 − i)
(d) i − 1.
Answer:
(b) 2i
a = 1 + i
On squaring both the sides, we get,
a2 = (1 + i)2
a2 = 1 + i2 + 2i
a2 = 11 + 2i (i2 = 1)
a2 = 2i
Page No 13.63:
Question 16:
If (x + iy)1/3 = a + ib, then
(a) 0
(b) 1
(c) −1
(d) none of these
Answer:
(d) none of these
Page No 13.63:
Question 17:
is equal to
(a)
(b)
(c)
(d) none of these.
Answer:
(b)
Page No 13.63:
Question 18:
The argument of is
(a) 60°
(b) 120°
(c) 210°
(d) 240°
Answer:
(d) 240°
Page No 13.63:
Question 19:
If , then z4 equals
(a) 1
(b) −1
(c) 0
(d) none of these
Answer:
(a) 1
Rationalising the denominator:
Page No 13.63:
Question 20:
If , then arg (z) equal
(a) 0
(b)
(c) π
(d) none of these.
Answer:
(a) 0
Page No 13.64:
Question 21:
(a)
(b)
(c)
(d) none of these
Answer:
(a)
Page No 13.64:
Question 22:
(a) 1
(b)
(c)
(d) none of these
Answer:
(b)
Page No 13.64:
Question 23:
(a)
(b)
(c)
(d)
Answer:
(c)
Page No 13.64:
Question 24:
If , then x2 + y2 =
(a) 0
(b) 1
(c) 100
(d) none of these
Answer:
(c) 100
Page No 13.64:
Question 25:
If , then Re (z) =
(a) 0
(b)
(c)
(d)
Answer:
(b)
Page No 13.64:
Question 26:
If then y =
(a) 9/85
(b) −9/85
(c) 53/85
(d) none of these
Answer:
(c)
Page No 13.64:
Question 27:
If , then =
(a) 1
(b) −1
(c) 0
(d) none of these
Answer:
(a) 1
Page No 13.64:
Question 28:
If θ is the amplitude of , than tan θ =
(a)
(b)
(c)
(d) none of these
Answer:
(b)
Page No 13.64:
Question 29:
If , then
(a)
(b)
(c) amp (z) =
(d) amp (z) =
Answer:
(d) amp (z) =
Page No 13.64:
Question 30:
The amplitude of is equal to
(a) 0
(b)
(c)
(d) π
Answer:
(c)
Page No 13.64:
Question 31:
The argument of is
(a)
(b)
(c)
(d)
Answer:
(a)
Page No 13.64:
Question 32:
The amplitude of is
(a)
(b)
(c)
(d)
Answer:
(c)
Page No 13.65:
Question 33:
The value of (i5 + i6 + i7 + i8 + i9) / (1 + i) is
(a)
(b)
(c) 1
(d)
Answer:
(a)
Page No 13.65:
Question 34:
equals
(a)
i
(b) −1
(c) −
i
(d) 4
Answer:
(c) i
Page No 13.65:
Question 35:
The value of is
(a) −1
(b) −2
(c) −3
(d) −4
Answer:
(b) 2
Page No 13.65:
Question 36:
The value of is
(a) 8
(b) 4
(c) −8
(d) −4
Answer:
(c) 8
Page No 13.65:
Question 37:
If lies in third quadrant, then also lies in third quadrant if
(a)
(b)
(c)
(d)
Answer:
Since, lies in third quadrant.
Now,
Since, also lies in third quadrant.
From (1) and (2),
Hence, the correct option is (c).
Page No 13.65:
Question 38:
If , where , then is
(a)
(b)
(c)
(d) none of these
Answer:
Since
,
Hence, the correct answer is option (a).
Page No 13.65:
Question 39:
A real value of x satisfies the equation
(a) 1
(b) −1
(c) 2
(d) −2
Answer:
Hence, the correct option is (a).
Page No 13.65:
Question 40:
The complex number z which satisfies the condition lies on
(a) circle x2 + y2 = 1
(b) the x−axis
(c) the y−axis
(d) the line x + y = 1
Answer:
Hence, the correct option is (b).
Page No 13.65:
Question 41:
If z is a complex number, then
(a)
(b)
(c)
(d)
Answer:
It is obvious that, for any complex number z,
Hence, the correct option is (b).
Page No 13.65:
Question 42:
Which of the following is correct for any two complex numbers z1 and z2?
(a)
(b)
(c)
(d)
Answer:
Since, it is known that
,
and
Hence, the correct option is (a).
Page No 13.65:
Question 43:
If the complex number satisfies the condition , then z lies on
(a) x−axis
(b) circle with centre (−1, 0) and radius 1
(c) y−axis
(d) none of these
Answer:
Hence, the correct option is (b).
Page No 13.65:
Question 44:
sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for
(a) x = nπ
(b)
(c) x = 0
(d) No value of x
Answer:
Given sin x + i cos 2x and cos x – i sin 2x are conjugate to each other
i.e sin x + i cos 2x = cos x – i sin 2x
i.e sin x – i cos 2x = cos x – i sin 2x
on comparing real and imaginary part,
sin x = cos x and cos 2x = sin 2x
i.e. sin x = cos x and 2cos2x – 1 = 2 sin x cos x
i.e 2cos2x – 1 = 2 cos x cos x (∴ sin x = cos x)
i.e 2cos2x – 1 = 2cos2x
i.e – 1 = 0
which is a false statement.
Hence no value of x exist
Therefore, the correct answer is option D.
Page No 13.66:
Question 45:
The real value of α for which the expression is purely real, is
(a)
(b)
(c) nπ
(d) none of these where n ∈ N.
Answer:
Given is purely real
Which is given to purely real
Hence, the correct answer is option C.
Page No 13.66:
Question 46:
The value of is equivalent to
(a) |z + 3|2
(b) |z – 3|
(c) z2 + 3
(d) none of these
Answer:
Since |z + 3|2
Hence, the correct answer is option A.
Page No 13.66:
Question 47:
If then n =
(a) 2m + 1
(b) 4m
(c) 2m
(d) 4m + 1 where m ∈ N
Answer:
Given :-
Hence, the correct answer is option B.
Page No 13.66:
Question 48:
The vector represented by the complex number 2 – i is rotated about the origin through an angle in the clockwise direction, the new position of point is
(a) 1 + 2i
(b) –1 –2i
(c) 2 + i
(d) –1 + 2i
Answer:
Given 2 – i is rotated by angle in the clockwise direction about the origin
Let Z' denote the new position and Z denote the previous p
Hence, the correct answer is option B.
Page No 13.66:
Question 49:
The real value of θ for which the expression is a real number, is
(a)
(b)
(c)
(d) none of theses
Answer:
Given is a real number
Since is a real number
Hence, the correct answer is option C.
Page No 13.66:
Question 50:
|z1 + z2| = |z1| + |z2| is possible if
(a)
(b)
(c) arg (z1) = arg (z2)
(d) |z1| = |z2|
Answer:
Since given |z1 + z2| = |z1| + |z2|
i.e |z1 + z2|2 = (|z1| + |z2|)2
Hence, the correct answer is option C.
Page No 13.66:
Question 51:
The equation |z + 1 – i| = |z – 1 + i| represents a
(a) straight line
(b) circle
(c) parabola
(d) hyperbola
Answer:
|z + 1 – i| = |z – 1 + i|
Let z = x + iy ; where x denote real part and y denote imaginary part of z
![straight i. straight e space open vertical bar x plus i y plus 1 minus i close vertical bar equals open vertical bar x plus i y minus 1 plus i close vertical bar
straight i. straight e space open vertical bar open parentheses x plus 1 close parentheses plus i open parentheses y minus 1 close parentheses close vertical bar equals open vertical bar open parentheses x minus 1 close parentheses plus i open parentheses 1 plus y close parentheses close vertical bar
Since space open vertical bar z close vertical bar equals square root of x squared plus y squared end root
straight i. straight e space open vertical bar open parentheses x plus 1 close parentheses plus i open parentheses y minus 1 close parentheses close vertical bar squared equals open vertical bar open parentheses x minus 1 close parentheses plus i open parentheses y plus 1 close parentheses close vertical bar squared
straight i. straight e space open parentheses x plus 1 close parentheses squared plus open parentheses y minus 1 close parentheses squared equals open parentheses x minus 1 close parentheses squared plus open parentheses y plus 1 close parentheses squared
straight i. straight e space x squared plus 2 x plus 1 plus y squared minus 2 y plus 1 space equals x squared minus 2 x plus 1 plus y squared plus 2 y plus 1
straight i. straight e space 4 x equals 4 y
straight i. straight e space x equals y semicolon space which space represents space straight a space straight space line](/img/site_content/ask-answer/loader.gif)
Hence, the correct answer is option A.
Page No 13.66:
Question 52:
The area of the triangle on the complex plane formed by the complex numbers z, –iz and z + iz is
(a) |z|2
(b)
(c)
(d) none of these
Answer:
For any complex number z, –iz represents complex number obtained by rotating z clockwise by angle.
Hence, z, –iz and z + iz represents a right angled triangle with sides z, –iz and hypotenus z + iz
∴ Area of triangle formed is
Hence, the correct answer is option C.
Page No 13.67:
Question 1:
The principal value of the argument of the complex number 1 –i is ____________.
Answer:
Since z = 1 – i = 1 + i(–1)
Since z lies in 4th quadrant.
∴argument is given by
Page No 13.67:
Question 2:
The polar form of (i25)3 is ____________.
Answer:
(i25)3
= (i24. i)3
Since i24 = 1
= (i)3 = i2. i = – i
i.e (i25)3 = – i
Here modulus and argument
Polar form of –i is
Page No 13.67:
Question 3:
The value of is ____________.
Answer:
Page No 13.67:
Question 4:
The complex number in polar form is ____________.
Answer:
Page No 13.67:
Question 5:
The sum of the series i + i2 + i3 +_____ upto 1000 terms is ____________.
Answer:
i +
i2 +
i3 +_____ 1000 terms
i.e
i +
i2 +
i3 +
i4 _____ +
i1000
= i + i2 – i + 1 + i5 _____ +
i1000
=
i – 1 –
i + 1 +
i5 + _______ +
i1000
= 0 +
i5 +
i6 + ________ +
i100
Similarly, sum of next form terms is also zero. (âµ 1000 = 4(250)
i.e multiple of 4
Hence,
i +
i2 +
i3 +_____
i1000 = 0
Page No 13.67:
Question 6:
The multiplicative inverse of (1 + i) is ____________.
Answer:
For z = 1 + i
Let us suppose multiplicative inverse of 1 + i is a + ib
then (1 + i) (a + ib) = 1
i.e a + ib + ai + i2b = 1
i.e a + ib + ia – b = 1
i.e a – b + i(a + b) = 1 + i0
On comparing, real and imaginary part, we get
a – b = 1 and a + b = 0
i.e a – b = 1 and a = – b
i.e a + a = 1
i.e multiplicative inverse of 1 + i is
Page No 13.67:
Question 7:
If |z| = 4 and then z = ____________.
Answer:
If |z| = 4 and
Page No 13.67:
Question 8:
If z1 and z2 are two complex numbers such that z1 + z2 is a real number, then z2 = ____________.
Answer:
Given for two complex numbers, z1 and z2, we have z1+ z2 is real number
Page No 13.67:
Question 9:
For any non-zero complex number z, arg (z) + arg = ____________.
Answer:
For complex number z
Say z = x + iy = reiθ where r = modulus of z, θ = argument of
Let us arg z = θ
Since arg = – arg z
⇒ arg z + arg = θ + (–θ)
i.e arg z + arg = 0
Page No 13.67:
Question 10:
If |z + 4| ≤ 3, then the greatest and least values of |z + 1| are _______ and ____________.
Answer:
Given |z + 4| ≤ 3
here |z + 1| = |z + 1 + 3 – 3| = |z + 4 + (–3)|
Since |a + b| ≤ |a| + |b| ≤ |z + 4| + |–3| = |z + 4| + 3
≤ 3 + 3 (given)
hence maximum value of |z + 1| is 6
|z + 1| = |z + 4 –3|
Since |a – b| ≥ ||a| – |b|| ≥ –|a| + |b|
⇒ |z + 1| ≥ – |z + 4| + 3
Since |z + 4| ≤ 3
⇒ –|z + 4| ≥ –3
i.e |z + 1| ≥ –|z + 4| + 3 ≥ –3 + 3 = 0
Hence, minimum value of |z + 1| is 0.
Page No 13.67:
Question 11:
The modulus and argument of are _______ and _______ respectively.
Answer:
Since complex number is
and argument is
∴ argument is given by
∴ modulus of and argument is .
Page No 13.67:
Question 12:
If then the locus of z is ____________.
Answer:
Given
which defines locus of a circle.
Page No 13.67:
Question 13:
If |z + 2i| = |z – 2i|, then the locus of z is ____________.
Answer:
Given |z + 2i| = |z – 2i|
for z = |x + iy|
|x + iy + 2i| = |x + iy – 2i|
Squaring both sides, |x + i(y + 2)|2 = |x + i (y – 2)|2
i.e x2 + (y + 2)|2 = x2 + (y – 2)|2
i.e x2 + y2 + 4 + 4y = x2 + y2 + 4 – 4y
i.e 8y = 0
i.e y = 0
∴ locus is perpendicular bisector of the segment joining (0, –2) and (0, 2)
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Question 14:
If |z + 2| = |z – 2|, then the locus of z is ____________.
Answer:
|z + 2| = |z – 2| for z = x + iy
i.e |x + iy + 2| = |x + iy – 2|
i.e |(x + 2) + iy| = |(x – 2) + iy|
Square both sides,
|(x + 2)| + iy|2 = |(x – 2)| + iy|2
i.e (x + 2)2 + y2 = (x – 2)2 + y2
i.e x2 + 4 + 4x + y2 = x2 + 4 – 4x + y2
i.e fx = 0
i.e x = 0
Hence, locus is perpendicular bisector of the segment joining (–2, 0) and (2, 0).
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Question 15:
If z = –1 +, then arg (z) = ____________.
Answer:
,
Since
z lies in IV quadrant.
⇒ argument of
z is π
.
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Question 16:
If x < 0 is a real number, then arg (x) = ____________.
Answer:
If x < 0
i.e z = x + i 0 and x is negative
Hence, z lies in II quadrant chg z = π.
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Question 17:
The real value of 'a' for which 3i3 – 2ai2 + (1 – a) i + 5 is real is ____________.
Answer:
3i3 – 2ai2 + (1 – a) i + 5
i.e 3i2 i – 2a(–1) + (1 – a) i + 5
i.e 3i + 2a + (1 – a) i + 5
i.e 2a + 5 + i(1 – a – 3)
i.e 2a + 5 + i (–a – 2)
Since the above expression is given to be real
⇒ –a – 2 = 0
⇒ a = – 2
Page No 13.67:
Question 18:
If |z| = 2 and arg (z) = , then z = ____________.
Answer:
for |z| = 2 = r arg z =
z = r (cos(arg z) + i sin (arg z))
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Question 19:
The value of where n ∈ N, is ____________.
Answer:
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Question 20:
The locus of z satisfying is ____________.
Answer:
Given and for z = x + iy
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Question 21:
The conjugate of the complex number is ____________.
Answer:
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Question 22:
If (2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = x + iy, then 5.8.13...(4 + n2) = ____________.
Answer:
Given:- (2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = x + iy
Taking modulus both sides, we get
|(2 + i) (2 + 2i) (2 + 3i) ...... (2 + ni) = |x + iy|
By squaring both sides, we get
Page No 13.67:
Question 23:
If the point representing a complex number lies in the third quadrant, then the point representing its conjugate lies in the ____________.
Answer:
If z lies in III quadrant.
i.e z = – x – iy ; x, y ≥ 0
i.e lies in II quadrant.
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Question 24:
The multiplication of a non-zero complex number by i rotates it through ____________ in the anti-clockwise direction.
Answer:
Let z = x + iy then iz = ix + i2y
i.e iz = ix – y
i.e iz = – y + ix
i.e iz is rotated by 90°.
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Question 25:
The complex number cosθ + i sinθ __________ be zero for any θ.
Answer:
z = cos
θ +
i sin
θ can never be zero for any
θ because
z = 0
⇒ cos
θ = 0 and sin
θ = 0
Since no such value of
θ exists
⇒ cos
θ +
i sin
θ can never be zero.
Page No 13.67:
Question 26:
The argument of the complex number is ____________.
Answer:
Since
z1 lies in II quadrant
Since
z2 lies in I quadrant
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Question 27:
If a complex number coincides with its conjugate, then it lies on ____________.
Answer:
Let z = x + iy and
Since z = (given)
⇒ x + iy = x – iy
⇒ iy = – iy
⇒ 2iy = 0
i.e y = 0
Then z lies an x-axis.
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Question 28:
The points representing the complex number z for which |z + 1| < |z – 1| lie on the left side of ____________.
Answer:
Given |z + 1| < |z – 1|
Let z = x + iy where x, y ∈R
Squaring both sides, we get
Hence, |
z + 1| < |
z – 1| lies on left side of
y-axis
Page No 13.68:
Question 29:
If three complex numbers z1, z2 and z3 are in A.P., then points representing them lie on ____________.
Answer:
Since z1, z2 and z3 are in A.P
Hence 2z2 = z1 + z3
i.e z2 is the mid-point of line joining z1 and z3
⇒ z1, z2 and z3 lie on a straight.
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Question 30:
The principal argument of i–1097 is ____________.
Answer:
Let z = i–1097
Hence, principle argument is .
Page No 13.68:
Question 31:
The value of is ____________.
Answer:
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Question 32:
If and then the point representing lies in ____________.
Answer:
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Question 33:
If 0 < arg (z) < π, then arg (z) – arg (–z) = ____________.
Answer:
For 0 < arg z < π
Let z = r(cosθ, i sinθ)
i.e arg z = θ
Then –z = – r(cosθ + i sinθ)
= – r (+ cosθ + i (+ sinθ))
= (–1) reiθ
= eiπreiθ
= rei(θ + π)
i.e arg (–z) =
θ + π
⇒ arg
z – arg(–
z) =
θ – θ – π
= – π
Page No 13.68:
Question 34:
For any two complex numbers z1, z2 and any real numbers a, b, |az1 – bz2|2 + |bz1 + az2|2 = ____________.
Answer:
For complex z1 and z2 and real numbers a and b
Page No 13.68:
Question 35:
Let z1 and z2 be two complex numbers such that |z1 + z2| = |z1| + |z2|, then arg (z1) – arg (z2) = ____________.
Answer:
Given for complex number z1 and z2
i.e angle between z1 and z2 is 0
i.e arg (z1) – arg z2 = 0
Page No 13.68:
Question 36:
Let z1 and z2 be two complex numbers such that |z1 + z2| = |z1 – z2|, then arg (z1) – arg (z2) = ____________.
Answer:
Given |z1 + z2| = | z1– z2|
On squaring both sides, |z1 + z2|2 = | z1– z2|2 (1)
i.e for
![z subscript 1 equals r subscript 1 open parentheses cos theta subscript 1 plus i space sin theta subscript 1 close parentheses
z subscript 2 equals r subscript 2 open parentheses cos theta subscript 2 plus i space sin theta subscript 2 close parentheses](/img/site_content/ask-answer/loader.gif)
Using identities
![open vertical bar z subscript 1 plus z subscript 2 close vertical bar squared equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared plus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses
open vertical bar z subscript 1 minus z subscript 2 close vertical bar squared equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared minus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses
therefore space from space left parenthesis 1 right parenthesis comma
straight i. straight e space open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared plus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses space equals open vertical bar z subscript 1 close vertical bar squared plus open vertical bar z subscript 2 close vertical bar squared minus 2 open vertical bar z subscript 1 close vertical bar open vertical bar z subscript 2 close vertical bar space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses
straight i. straight e space cos open parentheses theta subscript 1 minus theta subscript 2 close parentheses equals 0
rightwards double arrow theta subscript 1 minus theta subscript 2 equals straight pi over 2
straight i. straight e space arg space z subscript 1 minus arg space z subscript 2 equals straight pi over 2](/img/site_content/ask-answer/loader.gif)
Page No 13.68:
Question 37:
If |z1| = |z2| and arg then z1 + z2 = ____________.
Answer:
Let |z1| = |z2| = r
Let arg z1 = θ1 and arg z2 = θ2
Page No 13.68:
Question 1:
Write the values of the square root of i.
Answer:
Page No 13.68:
Question 2:
Write the values of the square root of −i.
Answer:
Page No 13.69:
Question 3:
If x + iy = , then write the value of (x2 + y2)2.
Answer:
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Question 4:
If π < θ < 2π and z = 1 + cos θ + i sin θ, then write the value of .
Answer:
Page No 13.69:
Question 5:
If n is any positive integer, write the value of .
Answer:
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Question 6:
Write the value of .
Answer:
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Question 7:
Write 1 − i in polar form.
Answer:
Page No 13.69:
Question 8:
Write −1 + i in polar form
Answer:
Page No 13.69:
Question 9:
Write the argument of −i.
Answer:
Page No 13.69:
Question 10:
Write the least positive integral value of n for which is real.
Answer:
Page No 13.69:
Question 11:
Find the principal argument of .
Answer:
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Question 12:
Find z, if
Answer:
We know that,
Thus, .
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Question 13:
If , then find the locus of z.
Answer:
Hence, the locus of
z is real axis.
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Question 14:
If , find the value of .
Answer:
Hence,
.
Page No 13.69:
Question 15:
Write the value of .
Answer:
Hence, .
Page No 13.69:
Question 16:
Write the sum of the series upto 1000 terms.
Answer:
We know that,
Thus, the sum of the series upto 1000 terms is 0.
Page No 13.69:
Question 17:
Write the value of .
Answer:
Let z be a complex number with argument θ.
Then,
⇒ argument of is −θ.
Thus, .
Page No 13.69:
Question 18:
If , then find the greatest and least values of .
Answer:
Hence, the greatest and least values of
is 6 and 0.
Page No 13.69:
Question 19:
For any two complex numbers z1 and z2 and any two real numbers a, b, find the value of .
Answer:
Hence,
.
Page No 13.69:
Question 20:
Write the conjugate of .
Answer:
∴ Conjugate of
Hence, Conjugate of
is
.
Page No 13.69:
Question 21:
If n ∈ , then find the value of .
Answer:
Thus, the value of
is 0.
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Question 22:
Find the real value of a for which is real.
Answer:
Since, is real.
Hence, the real value of a for which is real is −2.
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Question 23:
If , find z.
Answer:
We know that,
Hence, .
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Question 24:
Write the argument of .
Disclaimer: There is a misprinting in the question. It should be instead of .
Answer:
Let the argument of be α. Then,
Let the argument of be β. Then,
Let the argument of be γ. Then,
∴ The argument of
Hence, the argument of .
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