Question

Find the square root of $9+40i$

Answer 1

Given $z=9+40i$

$\left|z\right|=\sqrt{9^2+{40}^2}=\sqrt{81+1600}=\sqrt{1681}=41$

We have $Re\left(z\right)=9$ and $Im\left(z\right)=40>0$

$\because \sqrt{z}=\pm (\sqrt{\frac{\left|z\right|+Re\left(z\right)}{2}}+i\sqrt{\frac{\left|z\right|-Re\left(z\right)}{2}})$

Substituting the above values we get

$\sqrt{z}=\pm (\sqrt{\frac{41+9}{2}}+i\sqrt{\frac{41-9}{2}})$

$\Rightarrow \sqrt{z}=\pm (\sqrt{\frac{50}{2}}+i\sqrt{\frac{32}{2}})$

$\Rightarrow \sqrt{z}=\pm (\sqrt{25}+i\sqrt{16})$

$\therefore \sqrt{z}=\pm (5+4i)$