Given $z_1, z_2$ prove that $4z^2_1+9z^2_2 = 0$

Question

I need to show that given $z_1 = 9 + 9i$ and $z_2=6-6i$, $$4z_1^2+9z_2^2=0.$$

$$z_1 = 12.7(cos 45 + i sin 45)$$ $$z_2 = 8.5(cos 315 + i sin 315)$$

I changed the terms to polar form, applied De Moivre's Theorem, and got $4z^2_1+9z^2_2 = 50.8i-76.5i$, which is incorrect.

Answer 1

It is hard to know where is the mistake without seeing the computations. The problem is good: $z_1^2=81(1+i)^2=162i$, $z_2^2=36(1-i)^2=-72 i$, $4z_1^2+9z_2^2= 648i-648i=0$

Answer 2

In polar coordinates $z_1$ should be $$9+9i = \frac{18}{\sqrt{2}}\left(\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right) = \frac{18}{\sqrt{2}}\left(e^{\frac{i\pi}{4}}\right)$$ and $z_2$ should be $$6-6i = \frac{12}{\sqrt{2}}\left(\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}\right) = \frac{12}{\sqrt{2}}\left(e^{\frac{7i\pi}{4}}\right)$$ Does this agree with what you got? If not, try plugging these in and see if it works.

Answer 3

This problem (should be) pretty straightforward, and I don't think you need De Moivre's theorem, either. Sometimes it helps to factor our similar terms, as I did in the last step, to check your progress.

$z_1=9+9i, \space z_2=6-6i$

$$\implies z_1^2=81+81i+81i+81i^2 = 81-81+(2\cdot81)i = (2\cdot81)i,$$ $$z_2^2=36-36i-36i+36i^2 =36-36-(2\cdot36)i=-(9\cdot8)i.$$

$$4z_1^2+9z_2^2\implies4(2\cdot81)i-9(9\cdot8)i \implies 8\cdot81i-8\cdot81i =0. $$