Complex Numbers Questions Answers [SLST 2016]

In this post, we will discuss the questions from the previous year’s SLST exam on complex numbers with answers. The problems with solutions on complex numbers are given below.

Previous Years SLST Questions on Complex Analysis

Q1: If a+ib = ${\displaystyle \frac{c+i}{c-i}}$ where a, b,c are real numbers, then a²+b²=? [WB SLST 2016]

(A) (c²+1)/(c²-1) (B) -c²

(C) i (D) 1

Answer:

a+ib = ${\displaystyle \frac{c+i}{c-i}}$ = ${\displaystyle \frac{(c+i{)}^2}{(c-i)(c+i)}}$ = ${\displaystyle \frac{c^2-1+2ci}{c^2+1}}$

Thus, a=${\displaystyle \frac{c^2-1}{c^2+1}}$ and b=${\displaystyle \frac{2c}{c^2+1}}$

⇒ a²+b² = $({\displaystyle \frac{c^2-1}{c^2+1}}{)}^2$ + $({\displaystyle \frac{2c}{c^2+1}}{)}^2$

= ${\displaystyle \frac{(c^2-1{)}^2+4c^2}{(c^2+1{)}^2}}$

= ${\displaystyle \frac{(c^2+1{)}^2}{(c^2+1{)}^2}}$

= 1

So option(D) is correct.

Q2: For a complex number z, the number of solutions of the equation $z^2=\bar{z}$ is [WB SLST 2016]

(A) 1 (B) 3

(C) 4 (D) 2

Answer:

The given equation is $z^2=\bar{z}$

Taking modulus on both sides, we get that

$|z{|}^2=|z|$ as $|\bar{z}|=|z|$

⇒ |z| = 0,1

Case 1: |z|=0 implies that z=0.

Case 2: |z|=1. So the given equation becomes

z²= $\bar{z}$ = z^-1

⇒ z³=1

It has three solutions. So the total number of solutions of z²= z bar is equal to 4. So option(C) is correct.

Q3: If for a complex number z, |z|-z=1+2i, then the value of z is [WB SLST 2016]

(A) 2-$\frac{3}{2}$i (B) $\frac{3}{2}$-2i

(C) $\frac{3}{2}$+2i (D) $\frac{1}{2}$-2i

Answer:

Let x=x+iy. Then |z| = $\sqrt{x^2+y^2}$.

|z|-z=1+2i

⇒ $\sqrt{x^2+y^2}$ – x-iy =1+2i

Comparing the real and imaginary parts of both sides, we get that

$\sqrt{x^2+y^2}$ – x=1 and y=-2

⇒ $\sqrt{x^2+4}$ – x=1 as y=-2.

⇒ x²+4 =1+2x+x²

⇒ 2x+1=4

⇒ x=3/2

∴ z =x+iy = 3/2 -2i.

So option(B) is correct.

Q4: The modulus (r) and the amplitude (θ) of the complex number z=1+i tan $\frac{3\pi }{5}$ are [WB SLST 2016]

(A) r=sec $\frac{3\pi }{5}$, θ=$\frac{3\pi }{5}$ (B) r=-sec $\frac{3\pi }{5}$, θ=-$\frac{2\pi }{5}$

(C) r=-sec $\frac{3\pi }{5}$, θ=$\frac{2\pi }{5}$ (D) r=-sec $\frac{3\pi }{5}$, θ=-$\frac{3\pi }{5}$

Answer:

Let r(cosθ+isinθ) = 1+i tan $\frac{3\pi }{5}$.

Thus, rcosθ =1 and rsinθ=tan $\frac{3\pi }{5}$.

⇒ r² = 1+tan² $\frac{3\pi }{5}$ = sec² $\frac{3\pi }{5}$

⇒ r = -sec $\frac{3\pi }{5}$ as sec $\frac{3\pi }{5}$<0.

So we deduce that

cosθ =-cos $\frac{3\pi }{5}$ and sinθ =-sin$\frac{3\pi }{5}$

⇒ θ = π+$\frac{3\pi }{5}$. As θ>π, this is not the principal argument.

⇒ arg z = θ-2π = -2π/5.

So option(B) is correct.

Q5: Find the principal value of iⁱ. [WB SLST 2016]

Answer:

We have i = cos(π/2)+isin(π/2) = e^iπ/2.

Note that iⁱ can be written as

iⁱ = exp(i Log i)

= exp(i Log e^iπ/2)

= exp( i(iπ/2) )

= exp(-π/2)

So the principal value of iⁱ is equal to e^-π/2.

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Definite Integral Question Answers SLST 2016