Leonel Contreras

Answered

2022-07-01

Find an antiderivative for $\sqrt{z}$
I'm having some trouble on this homework problem:
If $\sqrt{z}$ is defined by $\sqrt{z}={e}^{\left(\mathrm{log}z\right)/2}$ for the branch of the log function defined by the condition $-\pi /2\le \mathrm{arg}\left(z\right)\le 3\pi /2$, 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 & Explanation

Rebekah Zimmerman

Expert

2022-07-02Added 32 answers

Step 1
We let our intuition tell us that we expect the antiderivative is $g\left(z\right)=\frac{2}{3}{z}^{\frac{3}{2}}$
We can write this in polar form and expand it to
$g\left(z\right)=\frac{2}{3}{e}^{\frac{3}{2}\mathrm{log}\left(r\right)}\mathrm{cos}\left(\frac{3}{2}\theta \right)+i\left(\frac{2}{3}{e}^{\frac{3}{2}\mathrm{log}\left(r\right)}\mathrm{sin}\left(\frac{3}{2}\theta \right)\right)=u\left(r,\theta \right)+iv\left(r,\theta \right)$
where $r=|z|$ and $\theta =\mathrm{arg}\left(z\right)$.
Then, taking partial derivatives yields
${u}_{r}={r}^{\frac{1}{2}}\mathrm{cos}\left(\frac{3}{2}\theta \right);{u}_{\theta }=-{r}^{\frac{3}{2}}\mathrm{sin}\left(\frac{3}{2}\theta \right);{v}_{r}={r}^{\frac{1}{2}}\mathrm{sin}\left(\frac{3}{2}\theta \right);{v}_{\theta }={r}^{\frac{3}{2}}\mathrm{cos}\left(\frac{3}{2}\theta \right)$
We can then easily verify that the Cauchy-Riemann equations hold, namely (for polar coordinates) ${u}_{r}=\frac{{v}_{\theta }}{r}$ and ${u}_{\theta }=-r{v}_{r}$.
Step 2
Since CR holds, we know that the function g(z) is analytic and has derivative given by ${g}^{\prime }\left(z\right)=\left[\mathrm{cos}\left(\theta \right){u}_{r}-\frac{\mathrm{sin}\left(\theta \right){u}_{\theta }}{r}\right]+i\left[\mathrm{cos}\left(\theta \right){v}_{r}-\frac{\mathrm{sin}\left(\theta \right){v}_{\theta }}{r}\right]$
Plugging in and following the algebra, we are left with ${g}^{\prime }\left(z\right)={r}^{\frac{1}{2}}{e}^{i\frac{\theta }{2}}=\sqrt{z}$. Thus, we've verified that g(z) is indeed an antiderivative for $\sqrt{z}$.
Next, we let $\gamma :\left[-1,1\right]\to \mathbb{C}$ be a path in the upper half-plane.
By the Fundamental Theorem of Calculus, we know ${\int }_{\gamma }f\left(z\right)dz=g\left(\gamma \left(b\right)\right)-g\left(\gamma \left(a\right)\right)$, where g is an antiderivative for f.
Applying this here, we get ${\int }_{\gamma }\sqrt{z}dz=\frac{2}{3}\left(1{\right)}^{\frac{3}{2}}-\frac{2}{3}\left(-1{\right)}^{\frac{3}{2}}=\frac{2}{3}-\frac{2}{3}\left({i}^{3}\right)=\frac{2}{3}+\frac{2}{3}i$

Kendrick Hampton

Expert

2022-07-03Added 7 answers

Step 1
Your problem would need to read $-\frac{\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.
Step 2
Your intuition is correct. To prove this, use the definition of derivative with respect to z for complex functions on $F\left(z\right)=\frac{2}{3}{z}^{\frac{3}{2}}$. 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 $\gamma \left(t\right)\in \mathbb{C}$ over [a,b] contained in the (open, connected) domain of definition for a holomorphic function F(z), we have $F\left(\gamma \left(B\right)\right)-F\left(\gamma \left(A\right)\right)={\int }_{a}^{b}{F}^{\prime }\left(\gamma \left(t\right)\right){\gamma }^{\prime }\left(t\right)dt={\int }_{\gamma }{F}^{\prime }\left(z\right)dz.$

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