Toronto Math Forum

Let $f(z) = e^2z$ and $g(z) = e^z$.

We have

$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$

So, $f(z) = 0 $ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)


$$f(z) = e^2z=0$$
$$z = 0$$

Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.

Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$

Then,

Call $h(z) = f(z) - g(z) = e^x - e^2 x$

Note that:

$$h(0) = 1 $$

$$h(1) = e - e^2 <0 $$

By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.

I.e. there is a REAL ROOT x where $0<x<1$.

We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.

Thus, within $|z| <1$ there is only one root, and that root is real.

The equation for the curve is defined to be

$$L(t) = 3i + 3e^{it}$$ for $\frac{\pi}{2} \geq t\geq \frac{-\pi}{2}$

Also, $L'(t)=3ie^{it}$

We're given that $f(z)=\overline{z}^2$ so all we have to do is parametrize the equation:

\begin{align*}

\int_{L}f(z)dz&=\int_{\pi/2}^{-\pi/2}(\overline{3i+3e^{it}})^2(3ie^{it})dt \\
&=3i\int_{\pi/2}^{-\pi/2}(-3i+e^{-it})^2(e^{it})dt \qquad\color{red}{\text{Misprint}}\\
&=3^3i\int_{\pi/2}^{-\pi/2}(i-e^{-it})^2(e^{it}) dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-1-2ie^{-it}+e^{-2it})(e^{it})dt \\
&=3^3i\int_{\pi/2}^{-\pi/2}(-e^{it}-2i+e^{-it})dt \\
&=-3^3i\int_{\pi/2}^{-\pi/2}(e^{it}-e^{-it}+2i)dt \\
&=-3^3i\left[\frac{1}{i}e^{it}+\frac{1}{i}e^{-it}+2it\right]^{-\pi/2}_{\pi/2} \\
&=\frac{-3^3i}{i}\left[e^{it}+e^{-it}+2i^2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3\left[e^{it}+e^{-it}-2t\right]^{-\pi/2}_{\pi/2} \\
&= -3^3[(e^{-i\pi/2}+e^{i\pi/2}-2(-\pi/2))-(e^{i\pi/2}+e^{-i\pi/2}-2(\pi/2))] \\
&= -3^3[(-i+i+\pi)-(i-i-\pi)] \\
&= -3^3[\pi+\pi] \\
&= -54\pi
\end{align*}

Let $f(u,v) = u + iv$ where $u = \cosh{x}\cos{y}$ and $v=\sinh{x}\sin{y}$

Note that

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &= \sinh{x}\cos{y}\\
\frac{\partial v}{\partial y} &= \sinh{x}\cos{y}\\
\frac{\partial u}{\partial y} &= -\cosh{x}\sin{y}\\
\frac{\partial v}{\partial x} &= \cosh{x}\sin{y}

\end{aligned}

\end{equation*}

Since

\begin{equation*}

\begin{aligned}

\frac{\partial u}{\partial x} &=  \frac{\partial v}{\partial y} \\

\frac{\partial u}{\partial y} &\neq -\frac{\partial v}{\partial x}

\end{aligned}

\end{equation*}

We know that $f$ is not analytic on $D$ by the contrapositve of Cauchy-Riemann Theorem.

So, $f$ is not sourceless or irrotational on $D$.

The error you are talking about is that the textbook used $dx$ instead of $\partial x$

Thus,  the proper equations should be

\begin{equation*}

\begin{aligned}

\int_{\gamma} u dx+v dy &= \iint_{\Omega} \left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right]dxdy=0 \\

\int_{\gamma} u dy - v dx &= \iint_{\Omega} \left[\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right]dxdy=2 \text{ area}(\Omega)

\end{aligned}

\end{equation*}