Prove $|e^{z^2+1}| \le e^2$

Question

Actually what I want to prove is $|\int_{D} e^{z^2+1} dz| \le 2\pi e^2$ for $D$ = the unit circle (centred at origin).

What I know is $|\int_\gamma f(z)dz| \le AB$ for $A$ the length of $\gamma$ (for whatever $\gamma$ you need eg smooth, continuously differentiable, simple/injective, Jordan, whatever) and $B \in \mathbb R$ assuming $|f(z)| \le B$ for all $z \in image(\gamma)$.

Then obviously $2\pi$ is the length. As for the $e^2$, I think of 2 ways.

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Way 1: We have $|z^2+1| \le |z|^2+1 \le 1+1=2$ and then somehow $|e^{z^2+1}| \le e^{|z^2+1|} \le e^2$.

I believe in general for any $f: G \to \mathbb C$ for any subset $G$ of $\mathbb C$, we have that $|e^{f(z)}| \le e^{|f(z)|}$ because for $f=u+iv$, we just do $|e^{f(z)}|=e^{u(z)}$ and $e^{|f(z)|}=e^{\sqrt{u^2(z)+v^2(z)}}$.

Question 1: Is this correct?

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Way 2: $$e^{z^2+1} = e^{x^2-y^2+1}e^{i(2xy)}$$

$$=e^{x^2-y^2+1} [\cos(2xy) + i\sin(2xy)]$$

Then

$$|e^{x^2-y^2+1} [\cos(2xy) + i\sin(2xy)]|$$

$$= |e^{x^2-y^2+1}| |\cos(2xy) + i\sin(2xy)|$$

$$= e^{x^2-y^2+1} |\cos(2xy) + i\sin(2xy)|$$

$$= e^{x^2-y^2+1} $$

$$= e^{x^2-y^2+1} \le e^2$$

because $x^2-y^2+1 \le 2$ because $x^2-y^2+1=x^2-(1-x^2)+1=2x^2 \le 2$ because $x^2 \le 1$ because $x \le 1$

Question 2: Is this correct?

Answer 1

What you wrote is correct. $$ |e^w| = e^{\operatorname{Re}(w)} \le e^{|w|} $$ holds for all complex numbers $w$, so $$ |e^{z^2+1}| \le e^{|z^2+1|} \le e^{|z|^2+1} = e^2 $$ for $z = 1$.

However, this is not needed to estimate the integral. According to Cauchy's integral theorem, $\int_{D} e^{z^2+1} dz$ is zero.

Answer 2

Both are correct.

Essentially, $e^z = e^{x+yi} = e^x (\cos(y) + i \sin(y))$, therefore $|e^z| = |e^x||\cos(y) + i \sin(y)| = e^x$.

That is, $|e^z| = e^{\text{Re}(z)}\le e^{|z|}$ as $\text{Re}(z)\le |z|$ and $e^x$ is a monotone function for real $x$. This justifies the step $|e^{z^2+1}|\le e^{|z^2+1}|$ in your first method.

Answer 3

Proof that the integral is 0 by Cauchy's theorem:

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Note: we're not yet allowed to say derivative of holomorphic are holomorphic or even that they are continuous. hell.

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Step 0: Parametrise $C_1(0)$ by $\gamma: [0,2\pi] \to \mathbb C, \gamma(t)=(\cos(t),\sin(t))$ s.t. $C_1(0)=image(\gamma)$.

Step 1: $f(z)=e^{z^2+1}$ is entire as a composition of 2 entire functions $e^{z}$ and $z^2+1$

Step 2: $\mathbb C$ is open in itself, connected and simply connected.

Step 3: $f'(z)=2ze^{z^2+1}= 2zf(z)$ is continuous on $\mathbb C$ because $f'(z)=2zf(z)$ is entire because each of the following are entire: $2z, f(z)$ ($f(z)$ is shown to be entire in Step 1) and because products of entire functions are entire.

Step 4: $\gamma$ is a piecewise smooth path since $\gamma$ is a smooth path since both its real and imaginary parts are smooth since each of the following are smooth on $2 \pi$: $\cos(t), \sin(t)$ (because they are smooth on $\mathbb R$) and because sums and products of smooth functions are smooth.

Step 5: $\gamma$ is closed because $\gamma(0)=(1,0)\stackrel{aka}{=}1+0i=1=\gamma(2 \pi)$.

Step 6: Therefore, by Steps 1-5, the integral is zero by the version of Cauchy's theorem below with $D = \mathbb C$.

• Let $f: D \to \mathbb C$ be analytic for open, connected and simply connected $D \subseteq \mathbb C$. Suppose $f'$ is continuous on $D$. Let $\gamma$ be a piecewise smooth closed path with image in $D$. Then $\int_{\gamma} f = 0$.

QED