# Equations with complex numbers

Could anyone help me to solve two equations with complex numbers? I would like to know if there is a way to solve them avoiding the usual substitution $z=a+ib$ because calculations are not very easy

$1)$$z^3\bar{z}+3z^2-4=0 I tried in this way z^2(z\bar{z}+3)=4 z^2(|z^2|+3)=4 But I'm stuck at this point 2)$$ \begin{cases} |z^2+1|=1 \\ 2Re(z)=|z^2| \end{cases}$

Here I tried to substitute $\omega=z^2$ but it doesn't work

Thanks a lot in advice

## 2 Answers

1) Your equation $z^2(z\bar{z}+3)=4$ brings us close to the end. Let $|z^2|=c$, and take the norm of both sides. Then $c(c+3)=4$, giving $c=1$.

2) This one is a little messier. There is the obvious solution $z=0$. We look for non-zero solutions. Expanding the first equation tells us that $$z^2\bar{z}^2+(z^2+\bar{z}^2)=0.$$ From the second equation, $z^2\bar{z}^2=4(\text{Re}(z))^2$. Letting $z=re^{i\theta}$ and cancelling the $r$ we get $$4\cos^2\theta+2\cos(2\theta)=0.$$ The rest is straightforward. Use $\cos(2\theta)=2\cos^2\theta-1$.

• Thanks! But why did you substitute also $z^2$ with $c$ ? Is in this case $|z^2(|z^2|+3)|=|z^2|(|z^2|+3)$ ? – Gianolepo Nov 21 '15 at 16:54
• Yes, the norm of a product is the product of the norms. – André Nicolas Nov 21 '15 at 16:58
• So we end up with the equation $z^2=1$. Not hard! – André Nicolas Nov 21 '15 at 17:15
• @FrancescoCaruso: I may be repeating a mistake, but it still cancels, giving $\cos\theta=\pm \frac{1}{2}$. But the second equation forces positive, and forces norm equal to $1$. So we get the two non-real cube roots of unity. Now in case you are concerned, substitute in the first equation. They work. – André Nicolas Nov 23 '15 at 16:55
• Oh, I see what your problem is, you are looking at the wrong equation. It is from $4(\text{Re}(z))^2+z^2+\bar{z}^2=0$ that I get the cancellation of $r^2$, and then trig equation. – André Nicolas Nov 23 '15 at 17:24

Actually, what you have done is pretty clever/lucky in that you know $(|z^2| + 3)$ is real so $z^2(|z^2| + 3) = 4$ means $z^2$ is real. So $z^2 = \pm |z^2|$. Set $w = z^2$ and you have either:

$w \ge 0; w^2 + 3w - 4 = 0 \implies (x + 4)(x - 1) \implies w = 1 \implies z = \pm 1$

$w < 0; w^2 - 3w - 4 = 0 \implies (x - 4)(x + 1) \implies w = -1 \implies z = \pm i$

So $z = \pm i, \pm 1$.