Solve equation for z: $z^2=3+4i ,(z=x+iy)$

Question

$$z^2=3+4i ,(z=x+iy)$$ Seems easy? That's what I thought!

I get a system of equations that I can't solve: $x^2-y^2=3$ and $2xiy=4i$ which I then get $x=\frac{2}{y}$

I can't solve the system of equations?

Plug the expression of $x$ in the first equation. Set $X=1/x^2$, you get a quadratic equation in $X$... — Jean Marie, Dec 18 '17 at 22:57

score 4 · Accepted Answer · answered Dec 18 '17 at 22:57

4

Hint

$$3+4i = 4+4i-1$$ $$ = 4+4i+i^2 $$

answered Dec 18 '17 at 22:57

WaveX

That's it! Thank you very much! Didn't really look at it this way – RiktasMath Dec 18 '17 at 23:01

score 2 · Answer 2 · answered Dec 18 '17 at 23:05

2

Taking norms, you have: $|z|^2 = |3 + 4i|$, so $x^2 + y^2 = 5$. Combining this with $x^2 - y^2 = 3$ yields $2x^2 = 8$, hence $x = \pm 2$.

answered Dec 18 '17 at 23:05

Alex Zorn

2 Answers2