Question

If z =$\frac{2-i}{i}$ = , then Re(z$^2$) + lm(z$^2$) is equal to

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (A)

1

Answer 2

$z=\frac{2-i}{i} \Rightarrow z^{2}= \frac{(2-i)^{2}}{i^{2}}$
$=\frac{4-4 i-1}{-1}=\frac{3-4 i}{-1}=4 i-3$
Now, $\operatorname{Re}\left(z^{2}\right) =-3$ and $\operatorname{lm}\left(z^{2}\right)=4$
$\therefore \operatorname{Re}\left(z^{2}\right)+\operatorname{lm}\left(z^{2}\right) =-3+4=1$