# Complex numbers, equation

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?

• Do you know the triangle inequality? May 19, 2023 at 6:01
• 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?
– user882145
May 19, 2023 at 6:03
• @MarkBennet Yes I do May 19, 2023 at 6:10
• @MartinR: It's a mistake, it shouldent be there. May 19, 2023 at 6:11
• Following up on the hint of @MarkBennet, do you know when equality holds in the triangle inequality in the complex plane? May 19, 2023 at 6:52

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