Here is the integral I want to evaluate $$ \int \mathbb{z}\sqrt{2\mathbb{z} + 5}\space dz$$ The question says that we have to restrict ourselves to the same branch of the square root. I am struggling to see how to factor that in this information.
I approach it in the following manner: $$ \int \mathbb{z}\sqrt{2\mathbb{z} + 5}\space dz$$
$$ =\frac{\mathbb{z}}{3}(2\mathbb{z}+5)^{\frac{3}{2}} -\int \frac{1}{3}(2\mathbb{z}+5)^{\frac{3}{2}}\space dz \space (Integrating\space By\space Parts)$$ $$= \frac{\mathbb{z}}{3}(2\mathbb{z}+5)^{\frac{3}{2}} - \frac{1}{15}(2\mathbb{z}+5)^{\frac{5}{2}} + C $$ $$ = \frac{(2\mathbb{z}+5)^{\frac{3}{2}}(3\mathbb{z}-5)}{15} + C$$
However, the given answer is $$\frac{1}{20}(2\mathbb{z}+5)^{\frac{5}{2}} - \frac{5}{6}(2\mathbb{z}+5)^{\frac{3}{2}} + C$$ which simplifies to $$ \frac{(2\mathbb{z}+5)^{\frac{3}{2}}(6\mathbb{z}-35)}{60} + C$$
What am I doing wrong?
The book's solution has a typo: Substituting $$u = 2 z + 5$$ transforms the integral to \begin{multline}\frac14 \int (u - 5) \sqrt{u} \,du = \frac14 \int \left(u^\frac32 - 5 u^\frac12\right)\,du \\ = \frac{1}{4} \left(\frac25 u^\frac52 - \frac{10}{3} u^\frac32\right) + C = \frac1{10} u^\frac52 - \frac56 u^\frac32 + C = \frac1{\color{red}{10}}(2 z + 5)^\frac52 - \frac56 (2 z + 5)^\frac32 + C, \end{multline} which agrees with your answer.
There's nothing special to the complex setting about this process.
The relevance of the branch cut for $\sqrt{\cdot}$ is that making a continuous choice of branch cut on some open set $U$ guarantees that we can make a well-behaved choice of antiderivative, rather than, say, needing different branch cuts of $(2 z + 5)^\frac52$ for different points in $U$.