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principal branch of log(1-i)
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if someone can go step by step i would really appreciate it
pv log(z) = Log(z) = ln|z| + i(Arg(z))
i have absolutely no clue on this problem
As in solutions to tan(x) = y, there are many values for x, but just one principal value (the one that solves Arctan(y) = x).
z=1-i |z| = sqrt(2) Arg(z) = -pi/4 |dw:1345525989165:dw|
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