Сomplex number equation: $z^3+\frac{(\sqrt2+\sqrt2i)^7}{i^{11}(-6+2\sqrt3i)^{13}}=0$

Question

please tell me how I can solve the following equation.

$$z^3+\frac{(\sqrt2+\sqrt2i)^7}{i^{11}(-6+2\sqrt3i)^{13}}=0$$

What formula should I use? If possible, tell me how to solve this equation or write where I can find a formula for solving such an equation. I searched for it on the Internet, but could not find anything useful.

Answer 1

$\sqrt2+\sqrt2i=2e^{\frac{\pi i}4}$.

And $-6+2\sqrt3i=4\sqrt3e^{\frac{\pi i}6}$.

And $i^{11}=-i$.

So we have $z^3-\frac{2^7e^{\frac{7\pi i}4}}{i\cdot (4\sqrt3)^{13}e^{\frac{13\pi i}6}}=0\implies z^3+\frac{i}{e^{\frac{5\pi i}{12}}2^{19}3^{\frac{13}2}}=0\implies z=-\frac1{576\cdot 2^{\frac13}\cdot 3^{\frac16}}e^{\frac{\pi i}{36}},-\frac{e^{\frac{\pi i}{36}}}{{576\cdot2^{\frac13}\cdot 3^\frac16}}\cdot e^{\frac{2\pi i}3}$ or $-\frac{e^{\frac{\pi i}{36}}}{{576\cdot2^{\frac13}\cdot 3^\frac16}}\cdot e^{\frac{4\pi i}3}$.

Answer 2

Simplify at first, $$(\sqrt{2}+\sqrt{2}i)^7=64\sqrt{2}(1-i)$$ to prove this write $$\sqrt{2}^7(1+i)^7$$ Calculate $$(1+i)^3\cdot (1+i)^3\cdot (1+i)$$

Answer 3

Hint

You have to use the exponential form of complex numbers: set $Z=r\mathrm e^{i\theta}$ and observe that, for instance, $$\sqrt 2+\sqrt 2i=\sqrt 2(1+i)=\sqrt 2\cdot \sqrt2\mathrm e^{i\tfrac\pi 4}=2\,\mathrm e^{i\tfrac\pi 4}, \;\text{&c.}$$ so that $\;(\sqrt 2+\sqrt 2i)^7=2^7 \mathrm e^{i\tfrac{7\pi}4}$.

Can you proceed similarly for the denominator?

Answer 4

Noticing that $\sqrt2+\sqrt2i=2\left(\frac{\sqrt2}2+i\frac{\sqrt2}2\right)$ and $-6+2\sqrt3i=4\sqrt3\left(-\frac{\sqrt3}2+i\frac12\right)$ should help you to express the given complex numbers in the polar form. You can then use de Moivre's formula to simplify the expression. From there you might be able to continue further.