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 = $\dfrac{c+i}{c-i}$ 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 = $\dfrac{c+i}{c-i}$ = $\dfrac{(c+i)^2}{(c-i)(c+i)}$ = $\dfrac{c^2-1+2ci}{c^2+1}$
Thus, a=$\dfrac{c^2-1}{c^2+1}$ and b=$\dfrac{2c}{c^2+1}$
⇒ a2+b2 = $(\dfrac{c^2-1}{c^2+1})^2$ + $(\dfrac{2c}{c^2+1})^2$
= $\dfrac{(c^2-1)^2+4c^2}{(c^2+1)^2}$
= $\dfrac{(c^2+1)^2}{(c^2+1)^2}$
= 1
So option(D) is correct.
Q2: For a complex number z, the number of solutions of the equation $z^2=\overline{z}$ is [WB SLST 2016]
(A) 1 (B) 3
(C) 4 (D) 2
Answer:
The given equation is $z^2=\overline{z}$
Taking modulus on both sides, we get that
$|z|^2=|z|$ as $|\overline{z}|=|z|$
⇒ |z| = 0,1
Case 1: |z|=0 implies that z=0.
Case 2: |z|=1. So the given equation becomes
z2= $\overline{z}$ = 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) = eiπ/2.
Note that ii can be written as
ii = exp(i Log i)
= exp(i Log eiπ/2)
= exp( i(iπ/2) )
= exp(-π/2)
So the principal value of ii is equal to e-π/2.
ALSO READ:
Definite Integral Question Answers SLST 2016