MAT334-2018F > Quiz-1

Q1: TUT 5101


Victor Ivrii:
Find all solutions of the given equation.
(z+1)^2 =1-i.

Mengyang Li:
Solution is scanned

Victor Ivrii:
Still, typed is better

Yatong Yu:
$(z+1)^2=1-i$ 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 $k=1$   $z=2^{1/4}e^{i7\pi  /8}-1=-2^{1/4}(\cos \pi  /8 - i\sin   \pi  /8)-1$

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