Question

For a complex number $z$ , if $z^{2}+\overset{-}{z}-z=4i$ and $z$ does not lie in the first quadrant, then (where $i^{2}=-1$ )

• A. $\left|z\right|=\sqrt{2}$
• B. $\left|z\right|=2\sqrt{2}$
• C. $arg \left(z\right)=\frac{- \pi }{4}$
• D. $arg \left(z\right)=\frac{3 \pi }{4}$

Answer 1

Answer: (A)

Put $z=x+i y$ $\Rightarrow x^{2}-y^{2}+2 i x y-2 i y=4 i$
$\Rightarrow \left\{\begin{array}{c}x^{2}-y^{2}=0 \Rightarrow x=y \text { or }-y \\ x y-y=2\end{array}\right.$
Case
(i) if $x=y,$ then $x^{2}-x-2=0$ $\Rightarrow x=-1,2$
$\Rightarrow \left\{\begin{array}{c}z_{1}=-1-i \\ z_{2}=2+2 i \text { (rejected) }\end{array}\right.$
Case (ii) if $x=-y,$ then $x^{2}-x+2=0$
$\Rightarrow x \in \phi$
Hence, $|z|=|-1-i|=\sqrt{2}$