SOLUTION: log (1 + cos x + i sin x)/(cos x – 1 – i sin x)

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Question 14373: log (1 + cos x + i sin x)/(cos x – 1 – i sin x)
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
log +%281+%2B+cos+x+%2B+i%2A+sin+x%29%2F%28cos+x+-+1+-+i%2A+sin+x%29++
Multiplying the complex conjugate of the denominator.
Since z = +%281+%2B+cos+x+%2B+i%2A+sin+x%29%2F%28cos+x+-+1+-+i%2A+sin+x%29++
=
=
=
= +%28sin+x+%28-+sin+x+%2B+i+cos+x+%29%29%2F+++%28cos+x+%28cos+x+-+1%29+%29+
= sin+x%2F%28cos+x+%28cos+x+-+1%29+%29+ %28-+sin+x+%2B+i+cos+x%29+
= +sin+x%2F%28cos+x+%281-cos+x%29+%29+ %28+sin+x+-+i+cos+x+%29+
= sin+x%2F%28cos+x+%281-cos+x+%29+%29 +%28cos%28x-+pi%2F2%29+%2B+i+sin+%28x-pi%2F2%29%29+
(polar coordinates) ...(***)
The definition of log z for complex number may be beyond your level.
Anyway,I give it here.
If the polar form for z = r e%5E%28i%2Atheta%29+,
Here, by (***) the given z =
(where the radius r = |sin x/(cos x (1-cos x ) )|,
the argument theta+=+x-pi%2F2+)
Hence, the natural logarithm ln(z).
ln(z) = ln(r) + i * theta = ln|sin x| - ln|cosx (1-cos x) + i %28x-pi%2F2%29
... Answer

This is a very boring question about calculations, try your best to understand the details.

Kenny