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### Messages - Yatong Yu

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##### Quiz-6 / Re: Q6 TUT 5101
« on: November 17, 2018, 04:19:12 PM »
let $f(z)=e^z-1$  $g(z)=e^{2z}-1$

$e^{2z}-1=0$ so $e^{2z}=1$  $e^z=\pm1$

when  $e^z=+1$
$f(z)=e^z-1=0$ $f'(z)=e^z=1\neq0$    so order=1
$g(z)=e^{2z}-1=0$ $g'(z)=2e^{2z}=2\neq0$    so order=1
1-1=0   removable

when  $e^z=-1$
$f(z)=e^z-1=-2\neq0$    so order=0
$g(z)=e^{2z}-1=0$ $g'(z)=2e^{2z}=2\neq0$    so order=1
1-0=1   simple pole

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##### Term Test 1 / Re: TT1 Problem 5 (night)
« on: October 19, 2018, 09:32:23 AM »
w= ez
∴w= e(x+yi)=ex∙eyi
=ex(cosy + isiny)
∴w = excosy + iexsiny
∴(excosy)2 +(iexsiny)2≥ 22
∴e2x(sin2y+cos2y)≥ 4
∴e2x≥4
∴x≥ln4/2=ln2
also excosy ≥ 0 => cos y ≥ 0 => π/2≥y≥0
exsiny≥ 0 =>siny ≥0 =>π ≥y ≥ 0
∴ {Z: Z = x + yi, x≥ ln2, π/2≥y≥0}

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##### Quiz-2 / Re: Q2 TUT 5101
« on: October 05, 2018, 11:05:13 PM »
f(z) = z³+(2i)³i/z+2i
= (z+2i)(z²-2iz+(2i)²)/z+2i
=z²-2iz-4
limz->2i f(z)=limz->2i z²-2iz-4
=(2i)²-2i(2i)-4
= -4+4-4
=-4

Practically  unreadable despite all insane html "mathematics".
\begin{aligned}f(z) &= \frac{z^3+(2i)^3i}{z+2i}\\ &= \frac{(z+2i)(z^2-2iz+(2i)^2)}{z+2i}\\ &=z^2-2iz-4\\ \lim_{z\to2i} f(z)&=\lim_{z\to2i} z^2-2iz-4\\ &=(2i)^2-2i(2i)-4\\ &= -4+4-4\\ &=-4 \end{aligned}

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##### Quiz-1 / Re: Q1: TUT 5101
« on: September 29, 2018, 05:58:01 PM »
$(z+1)^2=1-i$ which is $(1, -1)$ on the axis with angle $-\pi /4$ and length $2^{1/2}$ so $1-i = 2^{1/2}e^{i(-\pi /4+2k\pi )}$
$$z+1=(1-i)^{1/2} =\bigl(2^{1/2}e^{i(-\pi /4+2k\pi )}\bigr)^{1/2}$$ $$z=2^{1/4}e^{i(-\pi /8 + k\pi )}-1$$
when $k=0$   $z=2^{1/4}e^{-i\pi /8}-1=2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$
when $k=1$   $z=2^{1/4}e^{i7\pi /8}-1=-2^{1/4}(\cos \pi /8 - i\sin \pi /8)-1$

Do not use pitiful html options to display mathematical snippets! I fixed it

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