0
$\begingroup$

here is an exercise with its solution from the following book

EAN : $9782100727803$

Exercise:

Solve equation for unknown $z$ in $\mathbb{C}$ $$(E):\quad (z^{2} + 4^{z} + 1)^{2} + (3^{z} + 5)^{2}=0 $$

Solution: $$(E)\iff \begin{align} z^{2}+(4+3i)z+(1+5i)&=0\; (1)\\ \mbox{ Or }\hspace{4cm} &\mbox{}\\ z^{2}+(4-3i)z+(1-5i)&=0\; (2) \end{align}$$

Would someone explain to me why $z$ is a solution of $(1)$ iff its conjugate $\bar{z}$ is a solution of $(2)$

enter image description here

  • 1
    Take a complex conjugate of the first equation. – 2017-02-12
  • 1
    This is because complex conjugation is a field automorphism. – 2017-02-12
  • 0
    @bernard could you elaborate more please – 2017-02-12
  • 1
    Done, mylord! $$ – 2017-02-12

2 Answers 2

1

$(1)\iff (2)$ because complex conjugation is a field automorphism, which means:

  • it maps $\mathbf C$ to itself (this is the ἀυτός part);
  • denoting it with $c$, it is compatible with addition: $c(z+z')=c(z)+c(z')$;
  • and with multiplication: $c(zz')=c(z)c(z').$

As a consequence, , if $P$ is any polynomial with complex coefficients: $P(X)=c_0+c_1X+\dots+c_nX^n$, its conjugate is the polynomial $$\overline P(X)=\bar c_0+\bar c_1X+\dots+\bar c_nX^n,$$ and for any complex number $z$,we have $$\overline{P(z)}=\overline P(\bar z).$$ This formula also explains why the complex roots of a polynomial with real coefficients are pairwise conjugate, since in this case $P(X)=\overline P(X)$.

1

$$ 0 = z^{2}+(4+3i)z+(1+5i) $$ If you take the complex conjugate of both sides you get $$ \overline{0} = \overline{z^2} + \overline{(4+3i)} \overline{z} + \overline{1+5i} \iff \\ 0 = \overline{z}^2 + (4-3i) \overline{z} + 1-5i $$ thanks to $\overline{z_1 z_2} = \overline{z}_1 \overline{z}_2$ and $\overline{z_1 + z_2} = \overline{z}_1 + \overline{z}_2$.