# Determine the square roots of...

• Mar 5th 2012, 09:01 AM
Scoplex
Determine the square roots of...
DETERMINE THE SQUARE ROOTS OF 1-(√3)j

Help will be much appreciated.
• Mar 5th 2012, 03:44 PM
Soroban
Re: Determine the square roots of...
Hello, Scoplex!

Here's an algebraic approach . . .

Quote:

Determine the square roots of .$\displaystyle 1 - i\sqrt{3}$

$\displaystyle \text{Let }\,a + bi \:=\:\sqrt{1 - i\sqrt{3}}\;\text{ where }a\text{ and }b\text{ are real numbers.}$

$\displaystyle \text{Square: }\:(a + bi)^2 \:=\:1 - i\sqrt{3} \quad\Rightarrow\quad (a^2 - b^2) + 2abi \:=\:1 - i\sqrt{3}$

$\displaystyle \text{Equate real and imagiary components: }\:\begin{Bmatrix} a^2-b^2 \:=\:1 & [1] \\ 2ab \:=\:\text{-}\sqrt{3} & [2] \end{Bmatrix}$

$\displaystyle \text{From [2]: }\:b \:=\:\text{-}\frac{\sqrt{3}}{2a}\;\;[3]$

$\displaystyle \text{Substitute into [1]: }\:a^2 - \left(\text{-}\frac{\sqrt{3}}{2a}\right)^2 \:=\:1 \quad\Rightarrow\quad a^2 - \frac{3}{4a^2} \:=\:1$

. . . . $\displaystyle 4a^4 - 4a^2 - 3 \:=\:0 \quad\Rightarrow\quad (2a^2 + 1)(2a^2 - 3) \:=\:0$

$\displaystyle 2a^2 + 1 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:-\frac{1}{2} \quad \text{ but }a\text{ must be real.}$

$\displaystyle 2a^2 - 3 \:=\:0 \quad\Rightarrow\quad a^2 \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\pm\frac{\sqrt{3}}{\sqrt{2}}$

$\displaystyle \text{Substitute into [3]: }\:b \:=\:-\frac{\sqrt{3}}{2\left(\pm\frac{\sqrt{3}}{\sqrt{2} }\right)} \quad\Rightarrow\quad b \:=\:\mp\frac{1}{\sqrt{2}}$

$\displaystyle \text{Therefore: }\:a + bi \;=\;\pm\frac{\sqrt{3}}{\sqrt{2}} \mp\frac{1}{\sqrt{2}}i \;=\; \pm\left(\frac{\sqrt{3} - i}{\sqrt{2}}\right)$
• Mar 5th 2012, 04:30 PM
TheSaviour
Re: Determine the square roots of...
$\displaystyle 1-\sqrt{3}i = 2 \bigg(\cos{\frac{\pi}{3}}-i\sin{\frac{\pi}{3}}\bigg)$, thus by De Moivre

$\displaystyle \pm \sqrt{1-\sqrt{3}i} = \pm \sqrt{2}\left(\cos{\frac{\pi}{6}}-\sin\frac{\pi}{6}\right) = \pm \frac{1}{\sqrt{2}}(\sqrt{3}-i)$.