rvc
  • rvc
Please help. Separate the following complex number into real and imaginary parts.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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rvc
  • rvc
\[\log_{1-i}(1+i) \]
Jhannybean
  • Jhannybean
property of logs?
Michele_Laino
  • Michele_Laino
try to change the base please

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More answers

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

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