### Author Topic: Q3 TUT 0102  (Read 3060 times)

#### Victor Ivrii

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##### Q3 TUT 0102
« on: October 12, 2018, 06:13:33 PM »
Show that the function $w = g(z) = e^{z^2}$ maps the lines $\{x = y\}$ and $\{x = -y\}$ onto the circle $\{w\colon |w| = 1\}$.

Show  that $g$ maps each of the two pieces of the region $\{x + iy\colon x^2 > y^2\}$  onto the set $\{w\colon |w| > 1\}$ and each of the two pieces of the region $\{x + iy: x^2 < y^2\}$ onto the set $\{w\colon |w| < 1\}$.

Draw all regions and domains.
« Last Edit: October 14, 2018, 07:11:56 AM by Victor Ivrii »

#### Quentin King

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##### Re: Q3 TUT 0102
« Reply #1 on: October 12, 2018, 08:56:26 PM »
Let $z = x + iy$

Then $w$ can be rewritten as:

$w = e^{(x+iy)(x+iy)} \\ w = e^{(x^2-y^2+i2xy)} \\ w = e^{x^2-y^2}(\cos(2xy) + i\sin(2xy)) \\$

Now find the modulus of $w$

$|w| = |e^{x^2-y^2}\cos(2xy) + ie^{x^2-y^2}\sin(2xy))| \\ |w| = e^{x^2-y^2}\sqrt{\cos^2(2xy)+\sin^2(2xy)} \\ |w| = e^{x^2-y^2} \\$

- Now note that when you consider the line ${x=y}$:

$|w| = e^{x^2-x^2} \\ |w| = e^0 \\ |w| = 1$

- And similarly on the line ${x=-y}$:

$|w| = e^{(-y)^2-y^2} \\ |w| = e^0 \\ |w| = 1$

- For the region $\{x+iy : x^2 > y^2\}$:

$|w| = e^{x^2-y^2} > e^0 > 1$

- And finally for the region $\{x+iy : x^2 < y^2\}$:

$|w| = e^{x^2-y^2} < e^0 < 1$

EDIT: I dunno why, but my drawing attachment appears sideways to me in the preview. If you download it or right click and open the link in a new tab it looks okay though.
« Last Edit: October 14, 2018, 11:48:29 AM by Quentin King »

#### Victor Ivrii

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##### Re: Q3 TUT 0102
« Reply #2 on: October 14, 2018, 07:15:10 AM »
1) you need to escape sin, cos etc : \sin , \cos

2) Never use * as sign of multiplication ; in view of 1) you do not need it, but if you want you can use \cdot: $a\cdot b$, or \times, producing $a\times b$