Q3 TUT 0102

[Victor Ivrii]

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.

[Quentin King]

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.

[Victor Ivrii]

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 $a\times b$

Please correct your post