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# complex question

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Solve z^3 = 8*i in complex numbers.

Jun 18, 2021

#1
+2109
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(a+bi)^3 = 8i

-b^3*i - 3ab^2 + 3ab^2 * i + a^3

a^3 - 3ab^2 = 0

3ab^2 - b^3 = 8

Now I'm stuck.

Good luck, I hope you can figure it out.

Sorry I wasn't able to help more.

=^._.^=

Jun 19, 2021
#2
+26117
+2

Solve $$z^3 = 8*i$$ in complex numbers.

$$\begin{array}{|rcll|} \hline z^3 &=& 8*i \quad | \quad \text{cube root both sides} \\ z &=& \sqrt[3]{8i} \\ z &=& \sqrt[3]{8}\sqrt[3]{i} \\ \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline i &=& e^{i\frac{\pi}{2}} \\ \sqrt[3]{i}=i^{\frac{1}{3}} &=& \left(e^{i\left(\frac{\pi}{2}+ 2\pi*k \right)}\right)^{\frac13} \\ \mathbf{\sqrt[3]{i}} &=& \mathbf{e^{i\left(\frac{\pi}{6}+\frac{2\pi*k}{3}\right) } } \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline k=0: & \sqrt[3]{i} &=& e^{i \frac{\pi}{6} } = \cos(30^\circ)+i*\sin(30^\circ) \\ k=1: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{2\pi}{3}\right) } = \cos(150^\circ)+i*\sin(150^\circ) \\ k=2: & \sqrt[3]{i} &=& e^{i \left( \frac{\pi}{6}+\frac{4\pi}{3}\right) } = \cos(270^\circ)+i*\sin(270^\circ) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{2\sqrt[3]{i}} \\ \hline z_1 &=& 2*\left( \cos(30^\circ)+i*\sin(30^\circ) \right) \\ z_1 &=& \sqrt{3} + i \\\\ z_2 &=& 2*\left( \cos(150^\circ)+i*\sin(150^\circ) \right) \\ z_2 &=& -\sqrt{3} + i \\\\ z_3 &=& 2*\left( \cos(270^\circ)+i*\sin(270^\circ) \right) \\ z_3 &=& -2i \\ \hline \end{array}$$

Jun 19, 2021
edited by heureka  Jun 19, 2021
#3
+2109
0

Nice :))

Fancy latex too.

What strategy is that called?

I would like to learn it, it seems useful.

=^._.^=

catmg  Jun 19, 2021