Author Topic: TUT 0601 Quiz2  (Read 902 times)

Chengyin Ye

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TUT 0601 Quiz2
« on: January 30, 2020, 03:40:13 PM »
Question: Find the limit of the function at the given point or explain why it does not exist.
f(z)=(z3-8i)/(z+2i)
(zā‰ -2i) at z0=-2i
Solution:
f(z)=(z3-8i)/(z+2i)
f(z)=(z3+(2i)3)/(z+2i)
f(z)=((z+2i)(z2-2iz+(2i)2)/(z+2i)
f(z)=(z+2i)(z2-2iz+(2i)2
limz ā†’ -2i f(z) =limz ā†’ -2i (z+2i)(z2-2iz+(2i)2
=(-2i)2-2i(-2i)+(2i)2=12i2=-12
« Last Edit: January 30, 2020, 03:47:31 PM by Chengyin Ye »