• rvc

Please help. Separate the following complex number into real and imaginary parts.

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  • rvc

Please help. Separate the following complex number into real and imaginary parts.

Mathematics
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  • rvc
\[\log_{1-i}(1+i) \]
property of logs?
try to change the base please

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  • rvc
\[\frac{\log(1+i)}{log(1-i)}\]
here is the formula for base change: \[\Large {\log _{1 - i}}x = \frac{{{{\log }_e}x}}{{{{\log }_e}\left( {1 - i} \right)}}\]
if we apply my formula above, when x= 1+i, we will get the requested answer
  • rvc
? wait im confused
  • rvc
what formula?
  • rvc
please explain
  • rvc
yes i wrote the same above
the thing is its not separated into real and imaginary parts yet
\[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\] do the conjugate thingy next
  • rvc
|dw:1439121116009:dw|
  • rvc
can i apply that formula?
you can apply but then you'll need to do that conjugate thingy which will be complicated with your formula. instead try what ganeshie has suggested.
  • rvc
after the conversion into polar form what did ganeshie did further
know about "rationalizing the denominator" approach? same thing can be done to "real"ize the denominator. multiply numerator and denominator by conjugate of denominator
  • rvc
im not getting that @hartnn :(
recall the logarithm properties \[\log ab = \log a+\log b\] and \[\log a^b = b\log a\]
\[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r+\log e^{ix}}{\log r+\log e^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\] how about now ?
  • rvc
i did understand you step till\[\rm \frac{ log~re^{ix} }{ log~re^{-ix} }\]
  • rvc
your*
  • rvc
got it :) @ganeshie8
notice that \(\log \color{red}{r}e^{ix}\) is in form \(\log \color{red}{a}b\)
  • rvc
yep yep after that?
multiply numerator and denominator by log r + ix denominator will be of the form (p+iq)(p-iq) = p^2 +q^2
  • rvc
  • rvc
thanks all :)

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