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I have tried by substituting $z = a + bi$ to try and expand and solve this problem, but I only end up with two polynomials with not much progress to solving from there.

Edited: The two polynomials that I got are: $a^3 -5a^2 -a +ab^2 -5b^2 = 0$ and $b^3 -5b^2 + b +a^2b - 5a^2 = 0$. Not particularly helpful but I am not sure... can't seem to see past it. I have also attempted to use polar form but haven't gotten much progress in that either.

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  • $\begingroup$ You should edit your question to include the work you have done on the exercise. $\endgroup$ Sep 8 '20 at 19:06
  • $\begingroup$ If you end up with two polynomials, then show us the two polynomials. $\endgroup$
    – WhatsUp
    Sep 8 '20 at 19:06
  • $\begingroup$ Not sure if it will be meaningful but the two polynomials are: $a^3 -5a^2 -a +ab^2 -5b^2 = 0$ and $b^3 -5b^2 + b +a^2b - 5a^2 = 0$ $\endgroup$
    – IcyBloom
    Sep 8 '20 at 19:10
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$$\left(z+\frac{1}{z}\right)^2=(5+5i)^2+4=4+50i.$$

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Credit goes to Michael Rozenberg. I just had to write it down such that it is easier to see how the solution was obtained.

$$(z+1/z)^2 = z^2 + 2 + 1/z^2 = z^2 - 2 + 1/z^2 + 4$$ $$= (z^2 - 2 + 1/z^2) + 4= (z-1/z)^2 + 4 = (5+5i)^2 +4$$

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The hard way:

From the initial equation, $$z^2-5(1+i)z-1=0$$ we can draw $z$ (two values): $$z=\frac{5(1+i)\pm\sqrt{4+50i}}2$$

As the product of the roots is $-1$,

$$\frac1z=-\frac{5(1+i)\mp\sqrt{4+50i}}2$$

and

$$z+\frac1z=\pm\sqrt{4+50i}$$ from which the modulus is easy (and unique).

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