What is (e^{z*i})/(z^2*(z-1)) ?

The solution to (e^{z*i})/(z^2*(z-1)) is (cos(z))/(-z^2+z^3)+i(sin(z))/(-z^2+z^3)

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simplify (e^{zi})/(z^2(z-1))

simplify

\frac{e ^{z \cdot i}}{z ^{2} \cdot ( z - 1 )}

\frac{cos ( z )}{- z ^{2} + z ^{3}} + i \frac{sin ( z )}{- z ^{2} + z ^{3}}

Solution steps

\frac{e ^{z i}}{z ^{2} ( z - 1 )}

e^{z i} = cos (z) + i sin (z)

= \frac{cos ( z ) + i sin ( z )}{z ^{2} ( z - 1 )}

Rewrite

\frac{cos ( z ) + sin ( z ) i}{z ^{2} ( z - 1 )}

in standard complex form:

\frac{cos ( z )}{z ^{3} - z ^{2}} + \frac{sin ( z )}{z ^{3} - z ^{2}} i

= \frac{cos ( z )}{z ^{3} - z ^{2}} + \frac{sin ( z )}{z ^{3} - z ^{2}} i

cosh ((3 t - 6) U (t - 2) 3 t)

\frac{75 ( 4 ^{24} + 4 ^{23} \cdot 72 )}{4 ^{25}}

(a + b + c) (e + f + g) (h + i + j)

\frac{6 - x}{x ( x - 3 )}

\frac{3 x ( - 2 x - y + 3 ) ^{2}}{2}

What is (e^{z*i})/(z^2*(z-1)) ?
The solution to (e^{z*i})/(z^2*(z-1)) is (cos(z))/(-z^2+z^3)+i(sin(z))/(-z^2+z^3)