complex equation

qwesd.211

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Mar 10, 2022
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Calculate
(zˉ)4=4z\left ( \bar{z} \right )^4=4zI tried this, but i don't know if it's correct
(xiy)4=4(x+iy)(x-iy)^{4}=4(x+iy)(xiy)=(4x+4iy)14(x-iy)=(4x+4iy)^{\frac{1}{4}}r=42+42=32r=\sqrt{4^2+4^2}=\sqrt{32}θ=arctan(44)=π4\theta=\arctan (\frac{4}{4})=\frac{\pi}{4}k=0,w0=(32)14[cosπ4+2π(0))4+isinπ4+2π(0))4]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}]=(32)14[cosπ16+isinπ16]=1.512+3.008i=\left(\sqrt{32}\right)^\frac{1}{4}\left[\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right]=1.512+3.008iPlease help. :c
 
Calculate
(zˉ)4=4z\left ( \bar{z} \right )^4=4zI tried this, but i don't know if it's correct
(xiy)4=4(x+iy)(x-iy)^{4}=4(x+iy)(xiy)=(4x+4iy)14(x-iy)=(4x+4iy)^{\frac{1}{4}}r=42+42=32r=\sqrt{4^2+4^2}=\sqrt{32}θ=arctan(44)=π4\theta=\arctan (\frac{4}{4})=\frac{\pi}{4}k=0,w0=(32)14[cosπ4+2π(0))4+isinπ4+2π(0))4]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}]=(32)14[cosπ16+isinπ16]=1.512+3.008i=\left(\sqrt{32}\right)^\frac{1}{4}\left[\cos \frac{\pi}{16}+i\sin \frac{\pi}{16}\right]=1.512+3.008iPlease help. :c
It would be easier to start with
(zˉ)4=4z\left ( \bar{z} \right ) ^4 = 4z

(reiθ)4=4reiθ\left ( re^{-i \theta } \right ) ^4 = 4 r e^{i \theta }

r4e4iθ=4reiθr^4 e^{-4i \theta } = 4r e^{i \theta }

r3e5iθ=4r^3 e^{-5i \theta } = 4
and go from there.

-Dan
 
Here are a few often overlooked facts about complex numbers.
z=z|z|=|\overline{\,z\,}| and 1z=zz2\dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}.
With the fact that z=z|z|=|\overline{\,z\,}| we get 1z=zz2\dfrac{1}{\overline{\,z\,}}=\dfrac{z}{|z|^2}
 
It would be easier to start with
(zˉ)4=4z\left ( \bar{z} \right ) ^4 = 4z

(reiθ)4=4reiθ\left ( re^{-i \theta } \right ) ^4 = 4 r e^{i \theta }

r4e4iθ=4reiθr^4 e^{-4i \theta } = 4r e^{i \theta }

r3e5iθ=4r^3 e^{-5i \theta } = 4
and go from there.

-Dan
Sorry, i don't understand at all, i need to pass r3e5iθr^3 e^{-5i \theta } to "z" terms? :c
 
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