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Messages - Quentin King

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Quiz-3 / Re: Q3 TUT 0102
« 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.

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Quiz-2 / Re: Q2 TUT 0202
« on: October 05, 2018, 09:26:10 PM »
$(\frac{i+1}{\sqrt{2}})$ in polar coordinates is $\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})$

The roots of this equation are equally spaced on the unit circle around the origin, and the polar angle of $(\frac{i+1}{\sqrt{2}})^4$ is $\pi$

Therefore we know that $(\frac{i+1}{\sqrt{2}})^4 = \cos(\pi) + i\sin(\pi) = -1$

So finally, $\exp[\pi(\frac{i+1}{\sqrt{2}})^4] = \exp[-\pi]$

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