What is expand ((z-i)^4)/(z^8+1) ?

The solution to expand ((z-i)^4)/(z^8+1) is (1+z^4-6z^2)/(1+z^8)+i(4z-4z^3)/(1+z^8)

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expand ((z-i)^4)/(z^8+1)

expand

\frac{( z - i ) ^{4}}{z ^{8} + 1}

\frac{1 + z ^{4} - 6 z ^{2}}{1 + z ^{8}} + i \frac{4 z - 4 z ^{3}}{1 + z ^{8}}

Solution steps

\frac{( z - i ) ^{4}}{z ^{8} + 1}

(z - i)^{4} = z^{4} - 4 i z^{3} - 6 z^{2} + 4 i z + 1

= \frac{z ^{4} - 4 i z ^{3} - 6 z ^{2} + 4 i z + 1}{z ^{8} + 1}

Rewrite

\frac{z ^{4} - 4 i z ^{3} - 6 z ^{2} + 4 i z + 1}{z ^{8} + 1}

in standard complex form:

\frac{z ^{4} - 6 z ^{2} + 1}{z ^{8} + 1} + \frac{- 4 z ^{3} + 4 z}{z ^{8} + 1} i

= \frac{z ^{4} - 6 z ^{2} + 1}{z ^{8} + 1} + \frac{- 4 z ^{3} + 4 z}{z ^{8} + 1} i

\frac{\partial}{\partial x} ((z) (8 x yz - e^{z}))

lo g_{10} (x^{- 3} y^{2})

(2 x - 3 i)^{2}

(5 x + 4 + 14)^{2}

\frac{- 1}{3} 3^{2} + \frac{2}{3} 3 + \frac{8}{3}

What is expand ((z-i)^4)/(z^8+1) ?
The solution to expand ((z-i)^4)/(z^8+1) is (1+z^4-6z^2)/(1+z^8)+i(4z-4z^3)/(1+z^8)