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Let $f(z) = 3z^2 + 1$

Show that if $\vert z \vert = 1$ and $\vert f(z) \vert = 2$, then $z = \pm i$.

My attempt: Start with the equation $\vert f(z) \vert = 2$. So we can re-write $\vert f(z) \vert = \vert 3z^2 + 1 \vert$ as $2 = \vert 3z^2 + 1 \vert$.

But here Im stuck. I know that $i^2=-1$, but can I use that somehow?

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    $\begingroup$ Do you know the triangle inequality? $\endgroup$ May 19, 2023 at 6:01
  • $\begingroup$ If you've worked with the polar form for complex numbers, you are told that $ \ z \ = \ \cos \theta + i \sin \theta \ \ , \ $ since the modulus of $ \ z \ $ is $ \ 1 \ \ . \ $ How would you write $ \ 3z^2 + 1 \ $ in that form, and what would its modulus be? If that modulus equals $ \ 2 \ \ , \ $ what values will $ \ \theta \ $ have? $\endgroup$
    – user882145
    May 19, 2023 at 6:03
  • $\begingroup$ @MarkBennet Yes I do $\endgroup$
    – bestmate21
    May 19, 2023 at 6:10
  • $\begingroup$ @MartinR: It's a mistake, it shouldent be there. $\endgroup$
    – bestmate21
    May 19, 2023 at 6:11
  • $\begingroup$ Following up on the hint of @MarkBennet, do you know when equality holds in the triangle inequality in the complex plane? $\endgroup$ May 19, 2023 at 6:52

2 Answers 2

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Write $z=e^{i\theta}=\cos \theta+i \sin \theta$ and $z^{2}=\cos (2\theta)+i \sin (2\theta)$. $4=|f(z)|^{2}=9 \cos ^{2}(2\theta)+1+6\cos (2 \theta)+9 \sin ^{2}(2\theta)=10+6\cos (2 \theta)$. This gives $\cos (2 \theta)=-1$ ( and hence $\sin (2\theta)=1$). Thus, $z^{2}=-1$ and $z =\pm i$.

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$$ 4 = |3z^2+1|^2 = (3z^2+1)(\overline {3z^2}+1) = 9 |z|^4 + 6 \operatorname{Re} (z^2) + 1 = 10 + 6 \operatorname{Re} (z^2) $$ implies that $\operatorname{Re} (z^2) = -1$.

The only complex number on the unit circle with real part equal to $-1$ is $-1$. So $z^2 = -1$ and therefore $z=i$ or $z=-i$.

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