If z=x+iy, ∀ x,y∈ R, i2=-1, xy≠ 0 and |z|=2, then the imaginary part of (z + 2/z - 2) cannot be

Question

If z=x+iy, ∀ x,y∈ R, i2=-1, xy≠ 0 and |z|=2, then the imaginary part of (z + 2/z - 2) cannot be (A) 1 (B) 3 (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

Given, |z|=2⇒ x2+y2=4 &xy≠ 0 Let, z=x+iy=2(cos θ + i sin â¡ θ ) So, (z + 2/z - 2)=((x + 2) + i y/(x - 2) + i y) =((x2+y2-4)+i(y(x-2)-y(x+2))/(x-2)2+y2)=(-4 y/8-4 x) i So, the imaginary part =-(4 y/8 - 4 x)=(y/x - 2) =(- 2 sin θ /2 - 2 cos â¡ θ )=-cotâ¡(θ /2), θ ∈ (0 , 2 π ) Now, θ ≠ (π /2),π ,(3 π /2)⇒ cot (θ /2)≠ -1,0,1

Q. If $z = x + i y, \forall x, y \in R, i^{2} = - 1, x y \neq = 0$ and $∣ z ∣ = 2,$ then the imaginary part of $\frac{z + 2}{z - 2}$ cannot be

$1$

$3$

$2$

$4$

Q. If z=x+iy,∀x,y∈R,i2=−1,xy=0 and ∣z∣=2, then the imaginary part of z−2z+2​ cannot be

1

3

2

4

Q. If $z = x + i y, \forall x, y \in R, i^{2} = - 1, x y \neq = 0$ and $∣ z ∣ = 2,$ then the imaginary part of $\frac{z + 2}{z - 2}$ cannot be

$1$

$3$

$2$

$4$