Find trigonometric form of complex number $\frac{1}{2}-\frac{1}{\sqrt{2}}i$

Question

Find trigonometric form of complex number $\frac{1}{2}-\frac{1}{\sqrt{2}}i$.

I assume that this is related to trigonometric form of complex number

$$1 \pm \cos(\alpha) \pm i\sin(\alpha)$$ or similiar.

Is it possible to find such form without taking arctangent?

Answer 1

Just do it.

$|\frac 12 - \frac 1{\sqrt 2} i| = \sqrt {\frac 12^2 + \frac 12} = \frac {\sqrt 3} 2$

So $\frac 12 - \frac 1{\sqrt 2} i = \frac {\sqrt 3}2 (\frac 1{\sqrt 3} - \frac {\sqrt 2}{\sqrt 3}i)$

$\frac {1}{\sqrt 3}^2 + (-\frac {\sqrt 2}{\sqrt 3})^2 = 1$ so there exists a unique $\theta$ ($0 \le \theta < 2\pi$) so that $\cos \theta = \frac 1{\sqrt 3}$ and $\sin \theta = -\frac {\sqrt 2}{\sqrt 3}$.

In that case $\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac {-\frac {\sqrt 2}{\sqrt 3}}{\frac 1{\sqrt 3}} = -{\sqrt 2}$.

So $\theta$ would be $\arctan {-\sqrt 2}$ with some possible linear transformation to put it in the proper (4th) quadrant. As $\arctan$ returns values from $-\frac \pi 2$ to $ \frac \pi 2$ (the 4th and 1st quadrant) we are good.

$\theta = \arctan{-\sqrt 2} = -0.304087... \pi$. Which so far as I can tell has no rational interpretation.

So $ \frac 12 - \frac 1{\sqrt 2} i = \frac {\sqrt 3}2(\cos \arctan{-\sqrt 2} + \sin \arctan{-\sqrt 2} i) = \frac {\sqrt 3}2e^{i\arctan{-\sqrt 2}}$

Answer 2

You should write in form $$ z= |z|(\cos \alpha +i\sin \alpha)$$

The modulus $$|z| = \sqrt{{1\over 4}+{1\over 2}} = {\sqrt{3}\over2}$$

Since $\sin \alpha = -{\sqrt{2}\over \sqrt{3}}$ and $\cos \alpha = -{1\over 2\sqrt{3}}$ and you will get an $\alpha$ by solving this system.

Answer 3

HINT

You have $$ re^{it} = r\left(\cos t + i\sin t\right), $$ and note that $$ r^2 = \left(\frac12\right)^2+\left(\frac1{\sqrt2}\right)^2 = \frac34, $$ can you find $t$?

Answer 4

In general any complex number $z=x+iy$ can be represented as $re^{i\theta}$ where \begin{equation} r^2 = |z|^2 = x^2+y^2\end{equation} and \begin{equation} \theta = \tan^{-1} \Bigl(\frac{y}{x}\Bigr)\end{equation}

Here, \begin{equation} r = \sqrt{\frac{1}{4}+\frac{1}{2}} = \frac{\sqrt{3}}{2}\end{equation} \begin{equation} \theta = \tan^{-1}\Bigl(\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{2}}\Bigr) = \tan^{-1}(-\sqrt{2}) \end{equation} Here, since $ \cos\theta > 0 $ and $ \sin\theta <0, \theta $ is in fourth quadrant. Thus, $ \frac{3π}{2}\le\theta\le2π $ or simply, $\theta = -\tan^{-1} (\sqrt{2})$

Now assume that \begin{equation} \tan{x} = u \end{equation} Thus, \begin{equation} \cos = \sqrt{\frac{1}{1+u^2}} \end{equation} Thus, \begin{equation} x = \tan^{-1} (u) = \cos^{-1}\Bigl(\sqrt{\frac{1}{1+u^2}}\Bigr) \end{equation} Thus, \begin{equation} \cos(\tan^{-1} (u)) = \frac{1}{\sqrt{1+u^2}} \end{equation} Thus, \begin{equation} \cos(\tan^{-1}(-\sqrt{2})) = \frac{1}{\sqrt{3}}\end{equation} Thus,\begin{equation} \sin{\theta} =- \sqrt{\frac{2}{3}}\end{equation}