complex equation

[qwesd.211]

Calculate
[math]\left ( \bar{z} \right )^4=4z[/math]I tried this, but i don't know if it's correct
[math](x-iy)^{4}=4(x+iy)[/math][math](x-iy)=(4x+4iy)^{\frac{1}{4}}[/math][math]r=\sqrt{4^2+4^2}=\sqrt{32}[/math][math]\theta=\arctan (\frac{4}{4})=\frac{\pi}{4}[/math][math]k=0,w_0=(\sqrt{32})^\frac{1}{4} [\cos \frac{\frac{\pi}{4}+2\pi(0))}{4}+i\sin \frac{\frac{\pi}{4}+2\pi(0))}{4}][/math][math]=\left(\sqrt{32}\right)^\frac{1}{4}\left[\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right]=1.512+3.008i[/math]Please help. :c

 

[Harry_the_cat]

You can check if it is correct by substituting it back in to the original equation.

 

[topsquark]

Calculate
[math]\left ( \bar{z} \right )^4=4z[/math]I tried this, but i don't know if it's correct
[math](x-iy)^{4}=4(x+iy)[/math][math](x-iy)=(4x+4iy)^{\frac{1}{4}}[/math][math]r=\sqrt{4^2+4^2}=\sqrt{32}[/math][math]\theta=\arctan (\frac{4}{4})=\frac{\pi}{4}[/math][math]k=0,w_0=(\sqrt{32})^\frac{1}{4} [\cos \frac{\frac{\pi}{4}+2\pi(0))}{4}+i\sin \frac{\frac{\pi}{4}+2\pi(0))}{4}][/math][math]=\left(\sqrt{32}\right)^\frac{1}{4}\left[\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right]=1.512+3.008i[/math]Please help. :c

It would be easier to start with
[imath]\left ( \bar{z} \right ) ^4 = 4z[/imath]

[imath]\left ( re^{-i \theta } \right ) ^4 = 4 r e^{i \theta }[/imath]

[imath]r^4 e^{-4i \theta } = 4r e^{i \theta }[/imath]

[imath]r^3 e^{-5i \theta } = 4[/imath]
and go from there.

-Dan

 

[pka]

Here are a few often overlooked facts about complex numbers.
[imath]|z|=|\overline{\,z\,}|[/imath] and [imath]\dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}[/imath].
With the fact that [imath]|z|=|\overline{\,z\,}|[/imath] we get [imath]\dfrac{1}{\overline{\,z\,}}=\dfrac{z}{|z|^2}[/imath]
[imath][/imath]

 

[qwesd.211]

It would be easier to start with
[imath]\left ( \bar{z} \right ) ^4 = 4z[/imath]

[imath]\left ( re^{-i \theta } \right ) ^4 = 4 r e^{i \theta }[/imath]

[imath]r^4 e^{-4i \theta } = 4r e^{i \theta }[/imath]

[imath]r^3 e^{-5i \theta } = 4[/imath]
and go from there.

-Dan

Sorry, i don't understand at all, i need to pass [imath]r^3 e^{-5i \theta }[/imath] to "z" terms? :c