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Q3 TUT 0102
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Topic: Q3 TUT 0102 (Read 5781 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
»
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Quentin King
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Posts: 2
Karma: 8
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
»
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Victor Ivrii
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Posts: 2607
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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 $a\times b$
Please correct your post
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