I have to calculate the following log:
a) log(-4) b) log (3i)
I don't really know what to do..
a) $ log(-4) = log|-4| + i\cdot arg(-4) + 2ki\pi = log4 + ?? + 2ki\pi$
b) $ log(3i) = log|3i| + i\dot arg(3i) + 2ki\pi = log(3) + ?? + 2ki\pi$
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Do you know what the arg function is? – Henry Swanson Dec 08 '13 at 11:57
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Related : http://math.stackexchange.com/questions/201991/for-what-values-alpha-for-complex-z-lnz-alpha-alpha-lnz/202164#202164 – lab bhattacharjee Dec 08 '13 at 12:15
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Hi, I googled arg-function, it helped me alot. Thanks! – Vazrael Dec 08 '13 at 12:23
1 Answers
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HINT. Concerning the first question, you can note that
log(-4) = log(4 i^2) = log(4) + 2 log(e^(i Pi /2)) = log(4) + i Pi
Can you continue with this ?
Claude Leibovici
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