What is the integral of im(z-i) ?

The integral of im(z-i) is mz+(imz^2)/2+C

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integral of im(z-i)

\int im (z - i) d z

m z + \frac{im z ^{2}}{2} + C

Solution steps

\int im (z - i) d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= mi \cdot \int - i + z d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= mi (- \int i d z + \int z d z)

\int i d z = i z

\int z d z = \frac{z ^{2}}{2}

= mi (- i z + \frac{z ^{2}}{2})

Simplify

mi (- i z + \frac{z ^{2}}{2}) : m z + \frac{im z ^{2}}{2}

= m z + \frac{im z ^{2}}{2}

Add a constant to the solution

= m z + \frac{im z ^{2}}{2} + C

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