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Mathematics Question on Complex Numbers and Quadratic Equations

Let z≠−iz\ne-i be any complex number such that z−iz+i\frac{z-i}{z+i} is a purely imaginary number. Then z+1zz+\frac{1}{z} is :

A

0

B

any non-zero real number other than 1

C

any non-zero real number

D

a purely imaginary number

Answer

any non-zero real number

Explanation

Solution

Let z=x+iyz=x+iy
z−iz+i\frac{z-i}{z+i} is purely imaginary means its real part is zero.
x+iy−ix+iy+i=x+i(y−1)x+i(y+1)×x−i(y+1)x−i(y+1)\frac{x+iy-i}{x+iy+i}=\frac{x + i\left(y -1\right)}{x + i\left(y +1\right)}\times\frac{x - i\left(y +1\right)}{x - i\left(y +1\right)}
=x2−2ix(y+1)+xi(y−1)+y2−1x2+(y+1)2=\frac{x^{2}-2ix\left(y+1\right)+xi\left(y-1\right)+y^{2}-1}{x^{2}+\left(y+1\right)^{2}}
=x2+y2−1x2+(y+1)2−2xix2+(y+)2=\frac{x^{2}+y^{2}-1}{x^{2}+\left(y+1\right)^{2}}-\frac{2xi}{x^{2}+\left(y+\right)^{2}}
for pure imaginary, we have
x2+y2−1x2+(y+1)2=0\frac{x^{2}+y^{2}-1}{x^{2}+\left(y+1\right)^{2}}=0
⇒x2+y2=1⇒(x+iy)(x−iy)=1\Rightarrow x^{2}+y^{2}=1 \Rightarrow \left(x + iy\right) \left(x - iy\right) = 1
⇒x+iy=1x−iy=z\Rightarrow x+iy=\frac{1}{x-iy}=z
and 1z=x−iy\frac{1}{z}=x-iy
z+1z=(x+iy)+(x−iy)=2xz+\frac{1}{z}=\left(x+iy\right)+\left(x-iy\right)=2x
(z+1z)\left(z+\frac{1}{z}\right) is any non-zero real number