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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.

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

3 Answers 3

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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.

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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 $|y| < 1$.

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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

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