# Evaluate the integral $\int_{\gamma} e^{1/z}dz$

Consider the mapping on the complex plane given by $$w = e^{1/z}$$.

(a) What is the image of the set $$\{z : |z|<1\}$$?

(b) Sketch the image of the line $$y = x$$.

(c) Find a sequence of points in the pre-image of the point $$w = i$$ which converges to $$0$$.

(d) Evaluate the integral $$\int_{\gamma} e^{1/z}dz$$, where $$\gamma$$ is the positively oriented unit circle centered at the origin.

I tried to use the transformation $$w=1/z$$ and then the transformation $$w=e^z$$ in part (a) and (b), but I am not sure about the image of the latter transformation. I have no idea for part (c) and (d).

Could you please show me how to solve this problem? I really appreciate your help.

• for (d), it is a meromorphic integral on a circle, and you should have a formula for that. Jun 23, 2020 at 8:55
• for (a) you can prove that the transformations z -> 1/z maps { z : |z| < 1} to {z : |z| > 1} Jun 23, 2020 at 8:56

Observe that if $$\;z=a+bi\,,\,\,a,b\in\Bbb R\;$$ , then

$$\frac1z=\frac1{a+ib}=\frac{a-ib}{a^2+b^2}=\frac a{a^2+b^2}-\frac b{a^2+b^2}i$$

Thus, the map $$\;z\to\cfrac1z\;$$ changes the sign of the imaginary part. Also

$$|z|<1\implies\left|\frac1z\right|=\frac1{|z|}>1$$

and in fact

$$\left|e^{1/z}\right|=e^{a/(a^2+b^2)}$$

Thus, for example. with the line $$\;z=x+ix\;$$ . we get

$$\left|e^{1/z}\right|=e^{x/(x^2+x^2)}=e^{1/(2x)}$$

and choosing the argument to be in $$\;[0,2\pi)\;$$ , we get

$$e^{1/z}=i=e^{\pi i/2}\iff\frac1z=\frac\pi2+2k\pi=\frac\pi2\left(1+4k\right),\,\,k\in\Bbb Z\implies z=\ldots$$

And finally, using the power series of $$\;e^z\;$$ around zero, we get

$$e^{1/z}=\sum_{n=0}^\infty\frac1{n!z^n}=1+\frac1z+\frac1{2z^2}+\frac1{6z^3}+\ldots\implies \oint_\gamma e^{1/z}dz=\ldots$$

Fill in details, argue and finish the task.

This is on part (a).

The image is $$\mathbb{C} \setminus \{0\}$$; the image under $$1/z$$ is $$\{ z : |z| > 1 \}$$. Now we must show that the image of $$e^z$$ of $$\{ z : |z| > 1 \}$$ is $$\mathbb{C} \setminus \{0\}$$.

Now let $$w \neq 0$$. We can write $$w = e^{x + iy}$$. If $$|x + iy| > 1,$$ we are done.
If it happens that $$|x + iy| \leq 1,$$ notice that $$e^{x+i(y + 2\pi)} = w$$ too since the exponential has period $$2 \pi i,$$ and clearly $$|x+i(y + 2\pi)| \geq |y + 2\pi | > 1$$ since in this special case we must have $$|y| < 1$$.

For (a), just observe that the map $$z -> \frac{1}{z}$$ maps $${ z : |z| < 1}$$ to $${z : |z| > 1}$$

because if $$z = A e^{i \theta}$$ (so-called polar form).

then $$\frac{1}{z} = A^{-1}e^{- i \theta}$$

for (b) just calculate some points and sketch

for (c) you want $$e^{1/z}$$ to be $$i$$ which is equivalent to $$\frac{1}{z} = i * (\frac{\pi}{4} + 2 n \pi)$$ for some $$n \in \mathbb{N}$$

which is equivalent to $$z = \frac{1}{i * (\frac{\pi}{4} + 2 n \pi)}$$ which converges to 0 in n

for (d) If you have it you might want to use the residue theorem if you have it https://en.wikipedia.org/wiki/Residue_theorem