1
$\begingroup$

Question

if $z=2+4i$, find $\sqrt{z}$ when $\sqrt{z}=a+bi$

what I have so far

$\sqrt{z}=a+bi$, so I square both sides ->

$z=a^2+2abi-b^2$

we can substitute for z

$z=2+4i=a^2+2abi-b^2$

that means $a^2-b^2=2$ and $2abi=4i$

and ab=2 so then $b=\frac{a}{2}$

from then i dont know how to proceed from there

  • 3
    Substitute $b$ into $a^2-b^2=2$...and $b=\frac{2}{a}$... – 2017-02-03
  • 0
    An alternative way for finding square roots of complex numbers is expressing it in the form $re^{i\theta}$ because by demoivre's theorem $\sqrt{re^{i\theta}}=\sqrt{r}e^{\frac{i\theta}{2}}$ which can be done using half angle formulas – 2017-02-03

1 Answers 1

1

You have

$$a^2-b^2=2\quad (1)\\ ab=2\to b=\frac{2}{a}\quad (2)$$

Plug $(2)$ in $(1)$

$$a^2-\frac{4}{a^2}=2\to a^4-2a^2-4=0$$

so

$$a^2=\frac{2\pm\sqrt{20}}{2}=1\pm\sqrt{5}$$

so,

$$a=\pm\sqrt{1+\sqrt{5}}\to b=\pm\frac{2}{\sqrt{1+\sqrt{5}}}=\pm \frac{\sqrt{\sqrt{5}-1}}{2}$$