What is the limit as z approaches-i of (z^4-1)/(z+i) ?

The limit as z approaches-i of (z^4-1)/(z+i) is 4i

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limit as z approaches-i of (z^4-1)/(z+i)

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

4 i

Solution steps

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

Rationalize denominator

\frac{z ^{4} - 1}{z + i} : (z^{3} - z) + i (1 - z^{2})

= z \to - i lim (z^{3} - z + i (1 - z^{2}))

Plug in the value

z = - i

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

Simplify

(- i)^{3} - (- i) + i (1 - (- i)^{2}) : 4 i

= 4 i

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 4})

\int_{0}^{x} x cos (a x) d x

\frac{d y}{d x} + y = sin (x)

\int_{- \frac{π}{4}}^{0} 2 sec^{3} (x) d x

\frac{d}{d x} (1 - \frac{x}{2} - ln (2 + 9 x - x^{2}))

What is the limit as z approaches-i of (z^4-1)/(z+i) ?
The limit as z approaches-i of (z^4-1)/(z+i) is 4i