Thread: How to Computing Square root of a Complex number...

1. How to Computing Square root of a Complex number...

Hi,

How do I compute square root of a complexed number without a calculator?
In other words how should I think and break down the following with just pen and paper?

$\displaystyle\sqrt{8i}$

I would appriciate any guidance.
Thank you

2. Convert it to exponential form.

$\displaystyle \sqrt{8i} = \sqrt{8e^{\frac{\pi i}{2}}}$

$\displaystyle = 2\sqrt{2}(e^{\frac{\pi i}{2}})^{\frac{1}{2}}$

$\displaystyle = 2\sqrt{2}e^{\frac{\pi i}{4}}$

$\displaystyle = 2\sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right)$

$\displaystyle = 2\sqrt{2}\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}i}{2}\right)$

$\displaystyle = 2 + 2i$.

Note that this only gives one of the square roots, but you can find the other using the fact that there are always 2 square roots and they are evenly spaced around a circle.

3. Hello, 4Math!

Here is a primitive method . . .

Find: . $\sqrt{8i}$

Let: $a + bi \:=\:\sqrt{8i}$ . where $\,a$ and $\,b$ are real.

Square both sides: . $(a + bi)^2\;=\;(\sqrt{8i})^2$

And we have: . $(a^2-b^2) + (2ab)i \;=\;8i$

Equate real and imaginary components: . $\begin{array}{cccc}a^2-b^2 &=& 0 & [1] \\ 2ab &=& 8 & [2]\end{array}$

From [2] we have: . $b \:=\:\frac{4}{a}\;\;[3]$

Substitute into [1]: . $a^2 - \left(\tfrac{4}{a}\right)^2 \:=\:0 \quad\Rightarrow\quad a^4 \:=\:16$

. . Hence: . $a \:=\:\pm2$

Substitute into [3]: . $b \:=\:\dfrac{4}{\pm2} \:=\:\pm2$

Hence: . $a + bi \;=\;\pm2 \pm 2i$

Therefore: . $\sqrt{8i} \;=\;\pm2(1 + i)$