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?

  • 1
    $\begingroup$ Are you sure it's $e^{\color{red} {e\alpha}} $? $\endgroup$ – Rohan Jan 22 '18 at 14:21
  • $\begingroup$ @Rohan Good point, corrected. $\endgroup$ – Eenoku Jan 22 '18 at 14:23

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.


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 $e^{i\theta} = \cos \theta + i \sin \theta$.

  • $\begingroup$ 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 Jan 22 '18 at 14:25
  • $\begingroup$ See my edit...... $\endgroup$ – Ethan Bolker Jan 22 '18 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.