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How would one go about answering the following
Find all $z$ for which $$|z| = ze^{-i\frac \pi 2}$$

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closed as off-topic by Shailesh, Zain Patel, Misha Lavrov, Arnaud D., user91500 Apr 24 '17 at 6:04

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  • $\begingroup$ I presume you mean $e^{-i\pi/2}$. Hint: what is $e^{-i\pi/2}$? $\endgroup$ – Robert Israel Apr 24 '17 at 1:02
  • $\begingroup$ Notice that $e^{-i\pi/2}=-i,$ then $z=|z|i.$ $\endgroup$ – Sergio Charles Apr 24 '17 at 1:07
  • $\begingroup$ yeah i noticed that but i still don't see how to answer the question $\endgroup$ – David Abraham Apr 24 '17 at 2:01
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A perfectly good solution could start by evaluating $e^{-i\pi/2}$ to $-i$.

But, depending on which tools you already have, you could also start by dividing through by $e^{-i\pi/2}$ as it is, producing $$ z = |z|\cdot e^{\frac{\pi}2 i} $$ You will recognize the right-hand side of this as a complex number in polar form, so the equation states directly what the argument of $z$ needs to be.

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Method 1

Expand in Cartesian form $x+i y$: $$ \begin{align} % \left|z\right| &= z e^{-i\frac{\pi}{2}} \\\ % \sqrt{x^2+y^2} &= -i (x + i y) = y - ix % \end{align} $$ The left hand side is real, so the imaginary part of the right hand side must be $0$: $$ \boxed{ x = 0 } $$ Now the identity looks like $$ \sqrt{y^{2} = y} $$ which restricts $$ \boxed{ y>0 } $$


Method 2

Look for the roots of $$ f(z) = |z| - z e^{-i\frac{\pi}{2}} $$

Solve for the zeros of the real and imaginary parts $$ \text{Re } f = \sqrt{x^2+y^2}-y = 0 \qquad \Rightarrow \qquad \boxed{x = 0, y > 0} $$ $$ \text{Im } f = x = 0 \qquad \Rightarrow \qquad x = 0 $$

Check your work by quick computation of $f(i)$ and $f(1)$.


Method 3

The method of @Henning Makholm: polar form

$$ z = re^{i\theta} $$ $$ |z| = r, \qquad z e^{-i\frac{\theta}{2}} = re^{i\left(\theta-\frac{\theta}{2}\right)} $$ $$ \begin{align} |z| &= z e^{-i\frac{\pi}{2}} \\ r &= r e^{i\left(\theta-\frac{\pi}{2}\right)} \end{align} $$ When does $e^{i\left(\theta-\frac{\pi}{2}\right)} = 0$? When $\theta = \frac{\pi}{2}$, when we are on the positive imaginary axis.


Conclusion

The only numbers which satisfy the identity are purely imaginary. That is, Re $z = 0$.

Furthermore, this process maps points on the positive imaginary axis to points on equal magnitude on the positive real axis.

map

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    $\begingroup$ Hi, i get what you done here but does that really answer the 'find all z for' part. Surely z itself is complex? $\endgroup$ – David Abraham Apr 24 '17 at 3:05
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Another approach is geometric consideration. We know $e^{-i\frac{\pi}{2}}$ is a rotation about origin with $-\dfrac{\pi}{2}$ radiant and $|z|$ is a real number, so we can ask for what number in complex plane whose rotation by $-\dfrac{\pi}{2}$ will be a real number!

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