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### Messages - Xier Li

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##### Quiz-2 / Re: Q2 TUT 0201
« on: October 05, 2018, 08:27:59 PM »
$(z^4-1)/(z-i) = z^3+iz^2-z-i$ when $z\ne i$.
When $z\to i$, $z^4-1)/(z-i) \to -4i$.
This contradicts the fact that $f(z)=4i$ when $z=i$.
Thus, the function is continuous everywhere except $z=i.$

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##### Quiz-1 / Re: Q1: TUT 0201 and TU 0202
« on: September 28, 2018, 05:38:47 PM »
|z-(-1+2i)| = |z-(1-2i)|

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