Find an antiderivative for $\sqrt{z}$

Question

I'm having some trouble on this homework problem:

If $\sqrt{z}$ is defined by $\sqrt{z}=e^{(\log z)/2}$ for the branch of the log function defined by the condition $-\pi/2 \leq \arg(z) \leq 3\pi/2$, find an antiderivative for $\sqrt{z}$ and then find $\int_{\gamma} \sqrt{z} dz$, where $\gamma$ is any path from $-1$ to $1$ which lies in the upper half-plane.

I think the second part will be relatively straightforward by FTC once I find an antiderivative. What I'm having trouble with is the first part. My intuition from real-valued calculus says that an antiderivative for $\sqrt{z}$ is $2/3 z^{3/2}$. I'm not sure how to show this in the complex setting though. How do I use the log function and the associated branch to show this? Thanks.

Answer 1

Following Darrin's suggestion, we let our intuition tell us that we expect the antiderivative is $g(z)=\frac{2}{3}z^{\frac{3}{2}}$.

We can write this in polar form and expand it to

$$g(z)=\frac{2}{3}e^{\frac{3}{2}\log(r)}\cos\left(\frac{3}{2}\theta\right)+i\left(\frac{2}{3}e^{\frac{3}{2}\log(r)}\sin\left(\frac{3}{2}\theta\right)\right) = u(r, \theta)+iv(r,\theta)$$ where $r=|z|$ and $\theta = \arg(z)$.

Then, taking partial derivatives yields

$$u_r=r^{\frac{1}{2}}\cos\left(\frac{3}{2}\theta\right); u_{\theta}=-r^{\frac{3}{2}}\sin\left(\frac{3}{2}\theta\right); v_r = r^{\frac{1}{2}}\sin\left(\frac{3}{2}\theta\right); v_{\theta} = r^{\frac{3}{2}}\cos\left(\frac{3}{2}\theta\right)$$

We can then easily verify that the Cauchy-Riemann equations hold, namely (for polar coordinates) $u_r=\frac{v_{\theta}}{r}$ and $u_{\theta}=-rv_r$.

Since CR holds, we know that the function $g(z)$ is analytic and has derivative given by

$$g'(z)=\left[\cos(\theta)u_r - \frac{\sin(\theta)u_{\theta}}{r}\right]+i\left[\cos(\theta)v_r - \frac{\sin(\theta)v_{\theta}}{r}\right]$$

Plugging in and following the algebra, we are left with $g'(z)=r^{\frac{1}{2}}e^{i\frac{\theta}{2}} = \sqrt{z}$. Thus, we've verified that $g(z)$ is indeed an antiderivative for $\sqrt{z}$.

Next, we let $\gamma: [-1, 1] \to \mathbb{C}$ be a path in the upper half-plane.

By the Fundamental Theorem of Calculus, we know $\int_{\gamma} f(z)dz = g(\gamma(b))-g(\gamma(a))$, where $g$ is an antiderivative for $f$.

Applying this here, we get

$$\int_{\gamma} \sqrt{z} dz = \frac{2}{3}(1)^{\frac{3}{2}} - \frac{2}{3}(-1)^{\frac{3}{2}} = \frac{2}{3} -\frac{2}{3}(i^3) = \frac{2}{3} + \frac{2}{3}i$$

Answer 2

Your problem would need to read $-\frac{\pi}{2} < arg(z) < \frac{3\pi}{2}$, since including the possibility that one of the signs can be equal would mean $z^{\frac{1}{2}}$ is not continuous (and thus not holomorphic on the domain of definition), and including both possibilities for equality would mean that $z^{\frac{1}{2}}$ is not well-defined.

HINT: Your intuition is correct. To prove this, use the definition of derivative with respect to $z$ for complex functions on $F(z)=\frac{2}{3}z^{\frac{3}{2}}$. This will show that $F(z)$ is an antiderivative of $z^{\frac{1}{2}}$ on the domain of definition. The procedure works out identically to how the limit calculation would work in a calculus class.

To finish, note that $F(z)$ is itself holomorphic on the given domain of definition (why?). In such cases, it is true that for any $C^1$ path $\gamma(t) \in \mathbb{C}$ over $[a,b]$ contained in the (open, connected) domain of definition for a holomorphic function $F(z)$, we have $$F(\gamma(B))-F(\gamma(A)) = \int_a^b F'(\gamma(t))\gamma ' (t) dt=\int_{\gamma}F'(z)dz.$$