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I have a following equation

$$ |B| = \sqrt{\frac{2}{L}}, $$

$B$ being a complex number and $L$ being a real one.

The solution is supposed to be $$ B = \sqrt{\frac{2}{L}}e^{i\alpha}, $$

$\alpha$ being an arbitrary real number.

I can imagine the values being on a circle, all in the same distance from the point $B$, but I'm not able to derive the above-mentioned result mathematically, since it's been a pretty long time from my last complex analysis course.

I know, that: \begin{align} B &= x + iy\\ |B| &= \sqrt{x^2 + y^2}\\ \sqrt{x^2 + y^2} &= \sqrt{\frac{2}{L}}\\ x^2 + y^2 &= \frac{2}{L}, \end{align}

but this is obviously not the correct solution and I don't see the way to achieve the correct one step-by-step.

Could you help me?

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    $\begingroup$ Are you sure it's $e^{\color{red} {e\alpha}} $? $\endgroup$
    – user371838
    Commented Jan 22, 2018 at 14:21
  • $\begingroup$ @Rohan Good point, corrected. $\endgroup$
    – Eenoku
    Commented Jan 22, 2018 at 14:23

2 Answers 2

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These are just the points $B$ on the circle with radius $\sqrt{ 2/L}$ centered at the origin, so $$ B = \sqrt{\frac{2}{L}} e^{i \theta} $$ for arbitrary real $\theta$, which you can restrict to the interval $[0,2\pi)$.

You can reach this conclusion from your start by rewriting $x + iy$ in polar coordinates and using Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$.

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  • $\begingroup$ I can understand this, but is it possible to derive it step-by-step from the equation like I've attempted in the question? $\endgroup$
    – Eenoku
    Commented Jan 22, 2018 at 14:25
  • $\begingroup$ See my edit...... $\endgroup$ Commented Jan 22, 2018 at 14:28
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In general, given $z=x+iy$, it can be rewritten in polar form using $x=r\cos\theta$, $y=r\sin\theta$. Then $$z=r(\cos\theta+i\sin\theta)=re^{i\theta}$$where $r=|z|$, and $\theta=\arg z$.


You have some complex number $B$. This can be written as $$B=|B|e^{i\arg B}$$So, given that $|B|=\sqrt\frac2L$, this gives that $$B=\sqrt\frac2Le^{i\alpha}$$for $\alpha=\arg B$. Since no further information is given about $B$, we have that $\alpha$ is arbitrary, $B$ can have any argument.

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