Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz3 => Topic started by: Victor Ivrii 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.

Let $z = x + iy$
Then $w$ can be rewritten as:
$
w = e^{(x+iy)(x+iy)} \\
w = e^{(x^2y^2+i2xy)} \\
w = e^{x^2y^2}(\cos(2xy) + i\sin(2xy)) \\
$
Now find the modulus of $w$
$
w = e^{x^2y^2}\cos(2xy) + ie^{x^2y^2}\sin(2xy)) \\
w = e^{x^2y^2}\sqrt{\cos^2(2xy)+\sin^2(2xy)} \\
w = e^{x^2y^2} \\
$
 Now note that when you consider the line ${x=y}$:
$
w = e^{x^2x^2} \\
w = e^0 \\
w = 1
$
 And similarly on the line ${x=y}$:
$
w = e^{(y)^2y^2} \\
w = e^0 \\
w = 1
$
 For the region $\{x+iy : x^2 > y^2\}$:
$w = e^{x^2y^2} > e^0 > 1$
 And finally for the region $\{x+iy : x^2 < y^2\}$:
$w = e^{x^2y^2} < e^0 < 1$
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.

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