Question

Solve for $z:z^2-(3-2i)z=(5i-5)$.

• A. $z=(2+i)$ and $(1+3i)$
• B. $z=(2-i)$ and $(1-3i)$
• C. $z=(2+i)$ and $(1-3i)$
• D. $z=(2-i)$ and $(1+3i)$

Answer 1

The correct option is C $z=(2+i)$ and $(1-3i)$

$z^2-(3-2i)z=(5i-5)$
or $z^2-(3-2i)z-(5i-5)=0$
or $z=\frac{(3-2i)\pm \sqrt{(3-2i{)}^2+4(5i-5)}}{2}$
$=\frac{(3-2i)\pm \sqrt{9-4-12i-20}}{2}$
$=\frac{(3-2i)\pm \sqrt{8i-15}}{2}$
Now $\sqrt{-15+8i}$
$=\pm \{\sqrt{\frac{1}{2}\{\sqrt{(-15{)}^2+8^2}+(-15\left)\right\}}+i\sqrt{\frac{1}{2}\{\sqrt{(-15{)}^2+8^2}-(15\left)\right\}}\}$
$=\pm \{\sqrt{\frac{1}{2}(\sqrt{289}-15)}+\sqrt{\frac{1}{2}(\sqrt{289}+15)}\}$
$=\pm (1+i4)$
$⟹z=\frac{3-2i\pm (1+4i)}{2}$
$⟹z=(2+i)$ and $(1-3i)$
Ans: C