What is the integral from 0 to 1+i of z^2 ?

The integral from 0 to 1+i of z^2 is ((i+1)^3)/3

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integral from 0 to 1+i of z^2

\int_{0}^{1 + i} z^{2} d z

\frac{( i + 1 ) ^{3}}{3}

Solution steps

\int_{0}^{1 + i} z^{2} d z

Apply the Power Rule

: [\frac{z ^{3}}{3}]_{0}^{1 + i}

= [\frac{z ^{3}}{3}]_{0}^{1 + i}

Compute the boundaries:

\frac{( i + 1 ) ^{3}}{3}

= \frac{( i + 1 ) ^{3}}{3}

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What is the integral from 0 to 1+i of z^2 ?
The integral from 0 to 1+i of z^2 is ((i+1)^3)/3