Question

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

Answer 1

\[\log_{1-i}(1+i) \]

Answer 2

property of logs?

Answer 3

try to change the base please

Answer 4

\[\frac{\log(1+i)}{log(1-i)}\]

Answer 5

here is the formula for base change:
\[\Large {\log _{1 - i}}x = \frac{{{{\log }_e}x}}{{{{\log }_e}\left( {1 - i} \right)}}\]

Answer 6

if we apply my formula above, when x= 1+i, we will get the requested answer

Answer 7

? wait im confused

Answer 8

what formula?

Answer 9

please explain

Answer 10

yes i wrote the same above

Answer 11

the thing is its not separated into real and imaginary parts yet

Answer 12

\[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\]

do the conjugate thingy next

Answer 13

can i apply that formula?

Answer 14

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.

Answer 15

after the conversion into polar form what did ganeshie did further

Answer 16

know about "rationalizing the denominator" approach?

same thing can be done to "real"ize the denominator.

multiply numerator and denominator by conjugate of denominator

Answer 17

im not getting that @hartnn :(

Answer 18

recall the logarithm properties
\[\log ab = \log a+\log b\]
and
\[\log a^b = b\log a\]

Answer 19

\[\log_{1-i}(1+i) = \dfrac{\log re^{ix}}{\log re^{-ix}} = \dfrac{\log r+\log e^{ix}}{\log r+\log e^{-ix}} = \dfrac{\log r + ix}{\log r-ix}\]


how about now ?

Answer 20

i did understand you step till\[\rm \frac{ log~re^{ix} }{ log~re^{-ix} }\]

Answer 21

your*

Answer 22

got it :) @ganeshie8

Answer 23

notice that \(\log \color{red}{r}e^{ix}\) is in form \(\log \color{red}{a}b\)

Answer 24

yep yep
after that?

Answer 25

multiply numerator and denominator by log r + ix

denominator will be of the form (p+iq)(p-iq) = p^2 +q^2

Answer 26

thanks all :)