How to solve these complex equations?

Question

How to find z (complex number) in these equations? (i is an imaginary unit).

1) z + |z| = 2 + i

2) $2z^3 + \bar z^3 = 3$

I tried to find these with Oyler method but got stuck anyway. Huge thank you to anyone helping me out.

Answer 1

Hint

$$z=2-|z|+i$$

So, the imaginary part of $z$ is $1$

Consequently we have

$$a+i=2-\sqrt{a^2+1^2}+i$$ where $a$ is real

And $2-a=\sqrt{a^2+1}\ge1$

For the let $z^3=p+iq\implies\bar z^3=p-iq$

Answer 2

Let $z = a + bi; a,b \in \mathbb R$.

As $|z| = \sqrt{a^2 + b^2} \in \mathbb R$ we have

1) $z +|z| = 2+i$ so

$(a+\sqrt{a^2 + b^2}) + bi = 2+ i$ so

$b= 1$ and $a +\sqrt{a^2 + 1} = 2$.

So $\sqrt{a^2 + 1} = 2-a$

$a^2+1 = 4 - 4a + a^2$

$4a = 3$

$a =\frac 34$. and $z = \frac 34 + i$.

2) If $z = a+bi$ then $\overline{z} = a -bi$ and

$(a\pm bi)^3 = a^3 \pm 3a^2bi + 3ab^2i^2 \pm b^3i^3= (a^3-3ab^2) \pm(3a^2b - b^3)i$.

So $2z^3 +\overline{z}^3 = $

$3(a^3-3ab^2) +(3a^2b -b^3)i = 3$

So $3a^2b -b^3 = 0$ and $3(a^3 -3ab^2) = 3$

Solve those:

$3a^2b = b^3$ so either $b=0$ or $3a^2 = b^2$ and $b = \pm a\sqrt 3$

If $b = 0$ then $a^3 = 1$ and $a = 1$.

If $b = \pm a\sqrt3$ then $a^3- 3ab^2 = a^3 -9a^3 =1$ so $a=-\frac 12$.

So the solutions are either $z = 1$ and $2z^3 + \overline z = 2*1 + 1 =3$

Or $z = -\frac 12 \pm \frac {\sqrt 3}2i$.

But $z^3 = 1$ so $2z^3 +\overline z \not \in \mathbb R$ and this is not an acceptable solution.

So $z = 1$

Answer 3

For the second one note that $z^3+\bar z^3$ is real so $z^3$ is the difference between two real numbers. Hence $z^3$ is real and the equation can be reduced to a straightforward cubic with real coefficients and three solutions.