Complex number with z to the power of 4

Question

I have to find all $z\in C$ for which BOTH of the following is true:

1) $|z|=1$

2) $|z^4+1| = 1$

I understand that the 1) is a unit circle, but I can't find out what would be the 2). Calculating the power of $(x+yi)$ seems wrong because I get a function with square root of polynomial of the 8th degree.

I've even tried to equate the two, since their modulo is the same, like this:

$|z|= |z^4+1|$

and then separating two possibilities in which the equation holds, but I get only

$z^4\pm z+1=0$

which I don't know how to calculate.

I guess there is an elegant solution which I am unable to see...

Any help?

Answer 1

It seems helpful to square both sides. From the second equation, you get $$1 = (z^4 + 1)(\overline{z}^4 + 1) = (z \overline{z})^4 + z^4 + \overline{z}^4 + 1 = z^4 + \overline{z}^4 + 2.$$ That is, $z^4 + \overline{z^4} = -1.$

The first equation lets you write $z = e^{i \theta}$, so you have $$-1 = e^{4i\theta} + e^{-4i \theta} = 2\cos(4\theta);$$ in other words, $\cos(4\theta) = \frac{-1}{2}.$ The solutions of this are well-known.

Answer 2

Use polar coordinates to do part 2.

Setting $z=re^{i\phi}$ gives $z^4=r^4e^{i4\phi}.$ Therefore $$ z^4+1=(r^4\cos4\phi+1)+ir^4\sin4\phi, $$ and thus $$ |z^4+1|=1\Leftrightarrow (r^4\cos4\phi+1)^2+(r^4\sin4\phi)^2=1. $$ Expanding the terms on the l.h.s. of the last equation shows that the equation holds, iff $$ (r=0)\qquad \text{or}\qquad r^4+2\cos4\phi=0. $$ Therefore in polar form we get (in addition to the origin) the curve $$ r=\root 4\of{-2\cos4\phi}. $$ For the r.h.s. to be defined we need $\cos4\phi\le0$, so there will be something only in the sectors $\arg z\in[\pi/8,3\pi/8]$ and three other sectors gotten by rotating this with an integer multiple of $\pi/2$.

Below you see a Mathematica plot. The maximum modulus of $z$ is obviously $\root 4\of 2$, achieved at the points $z=\root 4\of 2e^{i(2k+1)\pi/4}$, $k=0,1,2,3,$ where we have $z^4=-2$.