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

1. rvc

$\log_{1-i}(1+i)$

2. Jhannybean

property of logs?

3. Michele_Laino

try to change the base please

4. rvc

$\frac{\log(1+i)}{log(1-i)}$

5. Michele_Laino

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

6. Michele_Laino

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

7. rvc

? wait im confused

8. rvc

what formula?

9. rvc

10. rvc

yes i wrote the same above

11. hartnn

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

12. 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

13. rvc

|dw:1439121116009:dw|

14. rvc

can i apply that formula?

15. 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.

16. rvc

after the conversion into polar form what did ganeshie did further

17. 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

18. rvc

im not getting that @hartnn :(

19. ganeshie8

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

20. 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 ?

21. rvc

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

22. rvc

your*

23. rvc

got it :) @ganeshie8

24. ganeshie8

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

25. rvc

yep yep after that?

26. hartnn

multiply numerator and denominator by log r + ix denominator will be of the form (p+iq)(p-iq) = p^2 +q^2

27. rvc

28. rvc

thanks all :)