Given
= (1 – i)n (1 + i)n
= (1 – i2)n
= 2n
Evaluate where n ϵ N.
Given
= (1 + i) (1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 + i13)
= (1 + i) i
= i + i2
= i – 1
If then find (x, y).
Given
= i3 – (-i3)
= 2i3
= 0 – 2i
Thus, (x, y) = (0, -2)
If the find the value of x + y.
Given
Rationalizing the denominator,
Thus,
Hence,
If then find (a, b).
Given
= (i4)25
= 1
∴ (a, b) = (1, 0)
If a = cos θ + i sin θ, find the value of
Given a = cos θ + i sin θ
∴ If a = cos θ + i sin θ, the value of
If (1 + i)z = (1 – i) then show that
Given (1 + i) z = (1 – i) z̅
Rationalizing the denominator,
= -iz̅
Hence proved.
If z = x + iy, then show that where bϵR, representing z in the complex plane is a circle.
Given z = x + iy
⇒ z̅ = x – iy
Now, z z̅ + 2 (z + z̅) + b = 0
⇒ (x + iy) (x – iy) + 2 (x + iy + x – iy) + b = 0
⇒ x2 + y2 + 4x + b = 0
This is the equation of a circle.
If the real part of is 4, then show that the locus of the point representing z in the complex plane is a circle.
Let z = x + iy
Now,
Given that real part is 4.
⇒ x2 + x – 2 + y2 = 4 (x2 – 2x + 1 + y2)
⇒ 3x2 + 3y2 – 9x + 6 = 0
Which represents a circle.
Hence, locus of z is a circle.
Show that the complex number z, satisfying the condition arg lies on a circle.
Let z = x + iy
Given
⇒ arg (z – 1) – arg (z + 1) = π/4
⇒ arg (x + iy – 1) – arg (x + iy + 1) = π/4
⇒ arg (x – 1 + iy) – arg (x + 1 + iy) = π/4
⇒ x2 + y2 – 1 = 2y
⇒ x2 + y2 – 2y – 1 = 0
Which represents a circle.
Solve that equation |z| = z + 1 + 2i.
Given |z| = z + 1 + 2i
Putting z = x + iy, we get
⇒ |x + iy| = x + iy + 1 + 2i
We know that
Comparing real and imaginary parts,
And 0 = y + 2
⇒ y = -2
Putting this value of y in ,
⇒ x2 + (-2)2 = (x + 1)2
⇒ x2 + 4 = x2 + 2x + 1
∴ x = 3/2
∴ z = x + iy
= 3/2 – 2i
If |z + 1| = z + 2 (1 + i), then find z.
Given |z + 1| = z + 2 (1 + i)
Putting z = x + iy, we get
⇒ |x + iy + 1| = x + iy + 2 (1 + i)
We know that
Comparing real and imaginary parts,
And 0 = y + 2
⇒ y = -2
Putting this value of y in ,
⇒ (x + 1)2 + (-2)2 = (x + 2)2
⇒ x2 + 2x + 1 + 4 = x2 + 4x + 4
⇒ 2x = 1
∴ x = 1/2
∴ z = x + iy
= 1/2 – 2i
If arg (z – 1) = arg (z + 3i), then find x – 1 : y. where z = x + iy
Let z = x + iy
Given arg (z – 1) = arg (z + 3i)
⇒ arg (x + iy – 1) = arg (x + iy + 3i)
⇒ arg (x – 1 + iy) = arg (x + I (y) = π/4
⇒ xy = xy – y + 3x – 3
⇒ 3x – 3 = y
∴ (x – 1): y = 1: 3
Show that represents a circle. Find its centre and radius.
Given
Substituting z = x + iy, we get
⇒ |x – 2 + iy| = 2 |x – 3 + iy|
⇒ x2 – 4x + 4 + y2 = 4 (x2 – 6x + 9 + y2)
⇒ 3x2 + 3y2 – 20x + 32 = 0
Hence, centre of circle is (10/3, 0) and radius is 2/3.
If is a purely imaginary number (z ≠ –1), then find the value of |z|.
Let z = x + iy
Now,
Given that is purely imaginary.
⇒ x2 – 1 + y2 = 0
⇒ x2 + y2 = 1
∴ |z| = 1
z1 and z2 are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that
Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)
Given that |z1| = |z2|
And arg (z1) + arg (z2) = π
⇒ θ1 + θ2 = π
⇒ θ1 = π – θ2
Now, z1 = |z2| (cos (π - θ2) + I sin (π - θ2))
⇒ z1 = |z2| (-cos θ2 + I sin θ2)
⇒ z1 = -|z2| (cos θ2 – I sin θ2)
⇒ z1 = - [|z2| (cos θ2 – I sin θ2)]
∴ z1 = -z̅ 2
Hence proved.
If |z1| = 1 (z1 ≠ –1) and then show that the real part of z2 is zero.
Let z1 = x + iy
Now,
Since x2 + y2 = 1
∴ Hence, the real part of z2 is zero.
If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find
Given z1 and z2 are conjugate complex numbers.
⇒ z2 = z̅ 1 = |z|/2pi is,
he negative side of the real axis
= r1 (cos θ1 - i sin θ1)
= r1 [cos (-θ1) + I sin (-θ1)]
Similarly, z3 = r2 (cos θ2 - i sin θ2)
⇒ z4 = r2 [cos (-θ2) + I sin (-θ2)]
= θ1 – (-θ2) + (-θ1) – θ2
= θ1 + θ2 – θ1 – θ2
= 0
If |z1| = |z2| = ….. = |zn| = 1, then
Show that |z1 + z2 + z3 + …. + zn|
Given |z1| = |z2| = … = |zn| = 1
⇒ |z1|2 = |z2|2 = … = |zn|2 = 1
⇒ z1 z̅ 1= z2 z̅ 2= z3 z̅ 3= … = zn z̅ n= 1
Now,
Hence proved.
If for complex numbers z1 and z2, arg (z1) – arg (z2) = 0, then show that |z1 – z2| = |z1| – |z2|.
Let z1 = |z1| (cos θ1 + I sin θ1) and z2 = |z2| (cos θ2 + I sin θ2)
Given arg (z1) - arg (z2) = 0
⇒ θ1 - θ2 = 0
⇒ θ1 = θ2
Now, z2 = |z2| (cos θ1 + I sin θ1)
⇒ z1 – z2 = ((|z1|cos θ1 - |z2| cos θ1) + i (|z1| sin θ1 - |z2| sin θ1))
We know that cos2 θ + sin2 θ = 1
∴ |z1 – z2| = |z1| - |z2|
Hence proved.
Solve the system of equations Re (z2) = 0, |z| = 2.
Given Re (z2) = 0, |z| = 2
Let z = x + iy.
Then
Given
⇒ x2 + y2 = 4 … (1)
Also, z2 = x2 + 2ixy – y2
= (x2 - y2) + 2ixy
Now, Re (z2) = 0
⇒ x2 – y2 = 0 … (2)
Solving (1) and (2), we get
⇒ x2 = y2 = 2
⇒ x = y = ±√2
∴ z = x + iy
= ±√2 ± i√2
Hence, we have four complex numbers.
Find the complex number satisfying the equation
Given equation z + √2 |(z + 1)| + i = 0 … (1)
Putting z = x + iy, we get
⇒ x + iy + √2 |x + iy + 1| + i = 0
Comparing real and imaginary parts to zero, we get
… (2)
And y + 1 = 0
⇒ y = -1
Putting y = -1 into (2), we get
⇒ 2x2 + 4x + 4 = x2
⇒ x2 + 4x + 4 = 0
⇒ (x + 2)2 = 0
⇒ x = -2
∴ z = x + iy
= -2 – i
Write the complex number in polar from.
Given
If z and w are two complex numbers such that |zw| = 1 and arg (z) – arg (w) = π/2, then show that
Let z = |z| (cos θ1 + I sin θ1) and w = |w| (cos θ2 + I sin θ2)
Given |zw| = |z| |w| = 1
Also arg (z) – arg (w) = π/2
⇒ θ1 - θ2 = π/2
Now, z̅ w = |z| (cos θ1 - I sin θ1) |w| (cos θ2 + I sin θ2)g | = 1
e get
ts to zero, we get
= |z| |w| (cos (-θ1) + I sin (-θ1)) (cos θ2 + I sin θ2)
= 1 [cos (θ2 – θ1) + I sin (θ2 – θ1)]
= [cos (-π/2) + I sin (-π/2)]
= 1 [0 – i]
= -i
Hence proved.
Fill in the blanks of the following
For any two complex numbers z1, z2 and any real numbers a, b, |az1 – bz2|2 + |bz1 + az2|2 = ….
Given |az1 – bz2|2 + |bz1 + az2|2
= |az1|2 + |bz2|2 – 2 Re (az1 × bz̅ 2) + |bz1|2 + |az2|2 + 2 Re (bz1 × az̅ 2)
= (a2 + b2) |z1|2 + (a2 + b2) |z2|2
= (a2 + b2) (|z1|2 + |z2|2)bz = |
Fill in the blanks of the following
The value of is ……………………………….
Given √-25 × √-9
= i √25 × i √9
= i2 (5 × 3)
= -15
Fill in the blanks of the following
The number is equal to ………………..
Given
= -2
Fill in the blanks of the following
The sum of the series i + i2 + i3 + …. Upto 1000 terms is…………
Given i + i2 + i3 + …. Upto 1000 terms
= (i + i2 + i3 + i4) + (i5 + i6 + i7 + i8) + … 250 brackets
Since in + in+1 + in + 2 + in + 3 = 0 where n ∈ N
= 0 + 0 + 0 … + 0
= 0
Fill in the blanks of the following
Multiplicative inverse of 1 + i is ………………..
Explanation:
Given 1 + i
= 1/2 (1 – i)
∴ Multiplication inverse of 1 + i is 1/2 (1 – i).
Fill in the blanks of the following
If z1 and z2 are complex numbers such that z1 + z2 is a real number, then z2 = ……..
Let z1 = x1 + iy1 and z2 = x2 + iy2
⇒ z1 + z2 = (x1 + x2) + i (y1 + y2) which is real
⇒ y1 + y2 = 0
⇒ y1 = - y2
Assuming x1 = x2
Since z2 = x1 – iy1
∴ z2 = z̅ 1
Fill in the blanks of the following
arg (z) + arg is …………….
⇒ arg (z) + arg z̅
= θ + (-θ)
= 0
Fill in the blanks of the following
If |z + 4| ≤ 3, then the greatest and least values of |z + 1| are ……… and …….
Given |z + 4| ≤ 3
For greatest value of |z + 1|,
⇒ |z + 1| = |z + 4 – 3|
⇒ |z + 1| ≤ |z + 4 – 3|
⇒ |z + 1| ≤ 3 + 3
⇒ |z + 1| ≤ 6
So, greatest value of |z + 1| is 6.
We know that least value of the modulus of a complex number is zero.
So, the least value of |z + 1| is zero.
Fill in the blanks of the following
If then the locus of z is ………….
Explanation:
Given
⇒ 6 |x – 2 + iy| = π |x + 2 + iy|
Squaring on both sides,
⇒ 36 |x – 2 + iy|2 = π2 |x + 2 + iy|2
⇒ 36 [x2 – 4x + 4 + y2] = π2 [x2 – 4x + 4 + y2]
⇒ (36 – π2) x2 + (36 – π2) y2 – (144 + 4π2) x + 144 – 4π2 = 0
Which is a circle.
Fill in the blanks of the following
If |z| = 4 and arg (z) = 5π/6, then z = ………..
Given |z| = 4 and arg (z) = 5π/6
Let z = |z| (cos θ + i sin θ) where θ = arg (z)
= -2√3 + 2i
∴ z = -2√3 + 2i
State True of False for the following:
The order relation is defined on the set of complex numbers.
False
Explanation:
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.
State True of False for the following:
Multiplication of a non zero complex number by –i rotates the point about origin through a right angle in the anti-clockwise direction.
False
Explanation:
Let z = x + iy, where x, y > 0
i.e. z or point A (x, y) lies in first quadrant.
Now, -iz = -I (x + iy)
= -ix – i2y
= y – ix
Now, point B (y, -x) lies in fourth quadrant.
Also ∠AOB = 90°
Thus, B is obtained by rotating A in clockwise direction about origin.
State True of False for the following:
For any complex number z the minimum value of |z| + |z – 1| is 1.
True
Explanation:
Given |z| + |z – 1|
We know that |z1| + |z2| ≥ |z1 – z2|
⇒ |z| + |z – 1| ≥ |z – (z – 1) |
⇒ |z| + |z – 1| ≥ 1
So, minimum value of |z| + |z – 1| is 1.
State True of False for the following:
The locus represented by |z – 1| = |z – i| is a line perpendicular to the join of (1, 0) and (0, 1).
True
Explanation:
Given |z – 1| = |z – i|
Putting z = x + iy,
⇒ |x – 1 + iy| = |x – i (1 – y) |
⇒ (x – 1)2 + y2 = x2 + (1 – y)2
⇒ x2 - 2x + 1 + y2 = x2 + 1 + y2 – 2y
⇒ -2x + 1 = 1 – 2y
⇒ -2x + 2y = 0
⇒ x – y = 0
Now, equation of a line through the points (1, 0) and (0, 1) is
⇒ x + y = 1
This line is perpendicular to the line x – y = 0.
State True of False for the following:
If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z2) = 0.
False
Explanation:
Given z ≠ 0 and Re (z) = 0
Let z = x + iy
Then x = 0
∴ z = iy
⇒ Im (z2) = i2y2 = -y2 ≠ 0
State True of False for the following:
The inequality |z – 4| < |z – 2| represents the region given by x > 3.
True
Explanation:
Given |z – 4| < |z – 2|
Putting z = x + iy,
⇒ |x – 4 + iy| < |x – 2 + iy|
⇒ (x – 4)2 + y2 < (x – 2)2 + y2
⇒ x2 – 8x + 16 + y2 < x2 – 4x + 4 + y2
⇒ -8x + 16 < -4x + 4
⇒ 4x > 12
∴ x > 3
State True of False for the following:
Let z1 and z2 be two complex numbers such that |z1 + z2| = |z1| + |z2|, then arg (z1 – z2) = 0.
False
Explanation:
Given |z1 + z2| = |z1| + |z2|
⇒ |z1 + z2|2 = |z1|2 + |z2|2 + 2 |z1| |z2|
⇒ |z1|2 + |z2|2 + 2 Re (z1z̅ 2) = |z1|2 + |z2|2 + 2 |z1| |z2|
⇒ 2 Re (z1 z̅ 2) = 2 |z1| |z2|
⇒ cos (θ1 – θ2) = 1
⇒ θ1 – θ2 = 0
∴ arg (z1) – arg (z2) = 0
State True of False for the following:
2 is not a complex number.
False
Explanation:
We know that any real number is also a complex number.
∴ 2 is a complex number.
Match the statements of Column A and Column B.
(a) Given z = i + √3
So, |z| = |i + √3|
= 2
Also, z lies in the first quadrant.
= π/6
∴ The polar form of z is
(b) Given z = -1 + √-3
= -1 + i √3
Here z lies in the second quadrant.
⇒ arg (z) = amp (z)
= π – tan-1 √3
= π - π/3
= 2π/3
(c) Given |z + 2| = |z – 2|
⇒ |x + 2 + iy| = |x – 2 + iy|
⇒ (x + 2)2 + y2 = (x – 2)2 + y2
⇒ x2 + 4x + 4 = x2 – 4x + 4
⇒ 8x = 0
∴ x = 0
It is a straight line which is a perpendicular bisector of segment joining the points (-2, 0) and (2, 0).
(d) Given |z + 2i| = |z – 2i|
⇒ |x + i(y + 2)| = |x + i(y - 2)|
⇒ (x)2 + (y + 2)2 = (x)2 + (y – 2)2
⇒ 4y = 0
∴ y = 0
It is a straight line which is a perpendicular bisector of segment joining the points (0, -2) and (0, 2).
(e) Given |z + 4i| ≥ 3
⇒ |x + iy + 4i| ≤ 3
⇒ |x + i(y + 4)|≤ 3
⇒ (x)2 + (y + 4)2 ≤ 9
⇒ x2 + y2 + 8y + 16 ≤ 9
⇒ x2 + y2 + 8y + 7 ≤ 0
This represents the region on or outside circle having centre (0, -4) and radius 3.
(f) Given |z + 4| ≤ 3
⇒ |x + iy + 4| ≤ 3
⇒ |x + 4 + iy|≤ 3
⇒ (x + 4)2 + y2 ≤ 9
⇒ x2 + 8x + 16 + y2 ≤ 9
⇒ x2 + 8x + y2 + 7 ≤ 0
This represents the region on or inside circle having centre (-4, 0) and radius 3.
(g) Given
Hence z̅ lies in the third quadrant.
(h) Given z̅ = 1 - i
lies in the third quadrant.
,outr inside circle having centre (-4, 0) and radius 3.nts (-2, x numbers is not possible or has no
= 1/2 (1 + i)
∴ Reciprocal of z lies in first quadrant.
What is the conjugate of
Explanation:
Given
Rationalizing the denominator,
If |z1| = |z2|, is it necessary that z1 = z2?
Explanation:
Given |z1| = |z2|
If |z1| = |z2| then z1 and z2 are at the same distance from origin.
But if arg (z1) ≠ arg (z2) then z1 and z2 are different.
So, if |z1| = |z2|, then it is not necessary that z1 = z2.
For example: z1 = 3 + 4i and z2 = 4 + 3i
Here |z1| = |z2| = 5 but z1 ≠ z2.
If what is the value of x2 + y2?
Explanation:
Given
Find z if |z| = 4 and arg (z) = 5π/6.
Explanation:
Given |z| = 4 and arg (z) = 5π/6
Let z = |z| (cos θ + i sin θ) where θ = arg (z)
= -2√3 + 2i
∴ z = -2√3 + 2i
Find
Explanation:
Given
= 1
Find principal argument of
Explanation:
Given z = (1 + i√3)2
= 1 – 3 + 2i√3
= -2 + i (2√3)
So, z lies in second quadrant.
= π – tan-1 √3
= π - π/3
= 2π/3
Where does z lie, if
Explanation:
Given
⇒ |z – 5i| = |z + 5i|
⇒ |x + iy – 5i| = |x + iy + 5i|
⇒ |x + i (y – 5) |2 = |x + i (y + 5) |2
⇒ x2 + (y – 5)2 = x2 + (y + 5)2
⇒ 20y = 0
⇒ y = 0
∴ z lies on the x – axis.
sinx + i cos2x and cos x – i sin 2x are conjugate to each other for:
A. x = nπ
B.
C. x = 0
D. No value of x
Given that sin x + i cos 2x and cos x – i sin 2x are conjugate to each other.
⇒ sin x - i cos 2x = cos x – i sin 2x
On comparing real and imaginary parts of both sides, we get
⇒ sin x = cos x and cos 2x = sin 2x
⇒ tan x = 1 and tan 2x = 1
Consider tan 2x = 1
We know that
This is not satisfied by tan x = 1.
Hence, no value of x is possible.
The real value of α for which the expression is purely real is:
A.
B.
C. nπ
D. None of these
Given
Rationalizing the denominator,
It is given that z is purely real.
⇒ -3 sin α = 0
⇒ sin α = 0
∴ α = nπ, n ∈ I
If z = x + iy lies in the third quadrant, the also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0
Since z = x + iy lies in the third quadrant, we get
X < 0 and y < 0
Now
Since also lies in third quadrant, we get
⇒ x2 – y2 < 0 and -2xy < 0
⇒ x2 < y2 and xy > 0
But x, y < 0
∴ y < x < 0
The value of (z + 3) is equivalent to
A. |z + 3|3
B. |z – 3|
C. z2 + 3
D. None of these
Given (z + 3) (z̅ + 3) = (x + iy + 3) (x –iy + 3)
= (x + 3)2 – (iy)2
= (x + 3)2 + y2
= |x + 3 + iy|2
= |z + 3|2
If , then
A. x = 2n + 1
B. x = 4n
C. x = 2n
D. x = 4n + 1
Given
Rationalizing the denominator,
⇒ ix = 1
∴ x = 4n, n ∈ N
A real value of x satisfies the equation if α2 + β2 =
A. 1
B. –1
C. 2
D. –2
Given
Rationalizing the denominator,
… (1)
… (2)
Multiplying equation (1) and (2),
So, α2 + β2 = 1
Which of the following is correct for any two complex numbers z1 and z2?
A. |z1 z2| = |z1||z2|
B. arg (z1z2) = arg (z1). Arg (z2)
C. |z1 + z2| = |z1|+ |z2|
D. |z1 + z2| ≥ |z1| – |z2|
Let z1 = |z1| (cos θ1 + i sin θ1) and z2 = |z2| (cos θ2 + i sin θ2)
Now, z1z2 = |z1| |z2| (cos θ1 + i sin θ1) (cos θ2 + i sin θ2)
= |z1| |z2| [cos θ1 cos θ2 + i sin θ1 cos θ2 + i cos θ1 sin θ2 + i2 sin θ1 sin θ2]
= |z1| |z2| [cos (θ1 + θ2) + i sin (θ1 + θ2)]
⇒ |z1 z2| = |z1| |z2|
And arg (z1 z2) = θ1 + θ2 = arg (z1) + arg (z2)
⇒ |z1 + z2| = |z1| + |z2| is true only when z1, z2 and O are collinear.
Also, |z1 + z2| ≥ ||z1| - |z2||
The point represented by the complex number 2 – i is rotated about origin through an angle π/2 in the clockwise direction, the new position of point is:
A. 1 + 2i
B. –1 – 2i
C. 2 + i
D. –1 + 2i
Given z = 2 – i
If z rotated through an angle of π/2 about the origin in clockwise direction.
Then the new position = z. e-(π/2)
= (2 – i) e-(π/2)
= (2 – i) [cos (-π/2) + i sin (-π/2)]
= (2 – i) (0 – i)
= -1 – 2i
Let x, y ϵ R, then x + iy is a non real complex number if:
A. x = 0
B. y = 0
C. x ≠ 0
D. y ≠ 0
X + yi is a non – real complex number if y ≠ 0 if x, y ? R.
If a + ib = c + id, then
A. a2 + c2 = 0
B. b2 + c2 = 0
C. b2 + d2 = 0
D. a2 + b2 = c2 + d2
Given a + ib = c + id
⇒ |a + ib| = |c + id|
Squaring on both sides,
∴ a2 + b2 = c2 + d2
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
Given
Let z = x + yi
⇒ |x + (y + 1) i| = |-x – (y – 1) i|
⇒ x2 + (y + 1)2 = x2 + (y – 1)2
⇒ (y + 1)2 = (y – 1)2
⇒ y2 + 2y + 1 = y2 – 2y + 1
⇒ 2y = -2y
⇒ 4y = 0
⇒ y = 0 (x – axis)
If z is a complex number, then
A. |z2| > |z|2
B. |z2| = |z|2
C. |z2| < |z|2
D. |z2| ≥ |z|2
Given z is a complex number.
Let z = x + yi
⇒ |z| = |x + yi| and |z|2 = |x + yi|2
⇒ |z|2 = x2 + y2 … (1)
Now z2 = x2 + y2i2 + 2xyi
⇒ z2 = x2 – y2 + 2xyi
So, |z|2 = x2 + y2 = |z|2
∴ |z|2 = |z2|
|z1 + z2| = |z1| + |z2| is possible if
A.
B.
C. arg (z1) = arg (z2)
D. |z1| = |z2|
Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)
Since |z1 + z2| = |z1| + |z2|
⇒ z1 + z2 = r1 cos θ1 + ir1 sin θ1+ r2 cos θ2 + ir2 sin θ2
But |z1 + z2| = |z1| + |z2|
Squaring both sides,
⇒ 2r1r2 – 2r1r2 cos (θ1 – θ2) = 0
⇒ 1 – cos (θ1 – θ2) = 0
⇒ cos (θ1 – θ2) = 1
⇒ (θ1 – θ2) = 0
⇒ θ1 = θ2
∴ arg (z1) = arg (z2)
The real value of θ for which the expression is a real number is:
A.
B.
C.
D. none of these
Let
Rationalizing the denominator,
If z is a real number,
⇒ 3 cos θ = 0
⇒ cos θ = 0
∴ θ = (2n + 1) π/2, n ∈ N
The value of arg (x) when x < 0 is:
A. 0
B.
C. π
D. none of these
Let z = x + 0i and x < 0
Since the point (-x, 0) lies on the negative side of the real axis,
∴ Principal argument (z) = π
If where z = 1 + 2i, then |f(z)| is
A.
B. |z|
C. 2|z|
D. none of these
Given where z = 1 + 2i
= |z|/2