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 $\sqrt{z}={e}^{(\mathrm{log}z)/2}$ for the branch of the log function defined by the condition $-\pi /2\le \mathrm{arg}(z)\le 3\pi /2$, 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.

I'm having some trouble on this homework problem:

If $\sqrt{z}$ is defined by $\sqrt{z}={e}^{(\mathrm{log}z)/2}$ for the branch of the log function defined by the condition $-\pi /2\le \mathrm{arg}(z)\le 3\pi /2$, 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 $g(z)=\frac{2}{3}{z}^{\frac{3}{2}}$

We can write this in polar form and expand it to

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

where $r=|z|$ and $\theta =\mathrm{arg}(z)$.

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 ${g}^{\prime}(z)=[\mathrm{cos}(\theta ){u}_{r}-\frac{\mathrm{sin}(\theta ){u}_{\theta}}{r}]+i[\mathrm{cos}(\theta ){v}_{r}-\frac{\mathrm{sin}(\theta ){v}_{\theta}}{r}]$

Plugging in and following the algebra, we are left with ${g}^{\prime}(z)={r}^{\frac{1}{2}}{e}^{i\frac{\theta}{2}}=\sqrt{z}$. 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 ${\int}_{\gamma}\sqrt{z}dz=\frac{2}{3}(1{)}^{\frac{3}{2}}-\frac{2}{3}(-1{)}^{\frac{3}{2}}=\frac{2}{3}-\frac{2}{3}({i}^{3})=\frac{2}{3}+\frac{2}{3}i$

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

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

where $r=|z|$ and $\theta =\mathrm{arg}(z)$.

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 ${g}^{\prime}(z)=[\mathrm{cos}(\theta ){u}_{r}-\frac{\mathrm{sin}(\theta ){u}_{\theta}}{r}]+i[\mathrm{cos}(\theta ){v}_{r}-\frac{\mathrm{sin}(\theta ){v}_{\theta}}{r}]$

Plugging in and following the algebra, we are left with ${g}^{\prime}(z)={r}^{\frac{1}{2}}{e}^{i\frac{\theta}{2}}=\sqrt{z}$. 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 ${\int}_{\gamma}\sqrt{z}dz=\frac{2}{3}(1{)}^{\frac{3}{2}}-\frac{2}{3}(-1{)}^{\frac{3}{2}}=\frac{2}{3}-\frac{2}{3}({i}^{3})=\frac{2}{3}+\frac{2}{3}i$

Kendrick Hampton

Expert

2022-07-03Added 7 answers

Step 1

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.

Step 2

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 $F(\gamma (B))-F(\gamma (A))={\int}_{a}^{b}{F}^{\prime}(\gamma (t)){\gamma}^{\prime}(t)dt={\int}_{\gamma}{F}^{\prime}(z)dz.$

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.

Step 2

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 $F(\gamma (B))-F(\gamma (A))={\int}_{a}^{b}{F}^{\prime}(\gamma (t)){\gamma}^{\prime}(t)dt={\int}_{\gamma}{F}^{\prime}(z)dz.$

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