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rvc
 one year ago
Please help.
Separate the following complex number into real and imaginary parts.
rvc
 one year ago
Please help. Separate the following complex number into real and imaginary parts.

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1try to change the base please

rvc
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\log(1+i)}{log(1i)}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is the formula for base change: \[\Large {\log _{1  i}}x = \frac{{{{\log }_e}x}}{{{{\log }_e}\left( {1  i} \right)}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if we apply my formula above, when x= 1+i, we will get the requested answer

rvc
 one year ago
Best ResponseYou've already chosen the best response.1yes i wrote the same above

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2the thing is its not separated into real and imaginary parts yet

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\log_{1i}(1+i) = \dfrac{\log re^{ix}}{\log re^{ix}} = \dfrac{\log r + ix}{\log rix}\] do the conjugate thingy next

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.1after the conversion into polar form what did ganeshie did further

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2know about "rationalizing the denominator" approach? same thing can be done to "real"ize the denominator. multiply numerator and denominator by conjugate of denominator

rvc
 one year ago
Best ResponseYou've already chosen the best response.1im not getting that @hartnn :(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2recall the logarithm properties \[\log ab = \log a+\log b\] and \[\log a^b = b\log a\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\log_{1i}(1+i) = \dfrac{\log re^{ix}}{\log re^{ix}} = \dfrac{\log r+\log e^{ix}}{\log r+\log e^{ix}} = \dfrac{\log r + ix}{\log rix}\] how about now ?

rvc
 one year ago
Best ResponseYou've already chosen the best response.1i did understand you step till\[\rm \frac{ log~re^{ix} }{ log~re^{ix} }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2notice that \(\log \color{red}{r}e^{ix}\) is in form \(\log \color{red}{a}b\)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2multiply numerator and denominator by log r + ix denominator will be of the form (p+iq)(piq) = p^2 +q^2
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