Hi, can someone please help me write $\displaystyle Log [(1+i)^4] $ in terms of its real and imaginary parts?
omigosh i am so lost in this subject
coz i thought that i had to use $\displaystyle Log z = ln|z| + i Arg(z) $
well this is what i did, but like yeahh i really dont know what im doing
$\displaystyle ln|1+i|^4 + i Arg(1+i)^4 $
$\displaystyle 4 ln|1+i| + i Arg(1+i)^4 $
umm please help me more sorry i didnt get how you got that answer
Convert to polars.
$\displaystyle x + iy = r(\cos{\theta} + i\sin{\theta}) = r\,e^{i\theta}$
You have $\displaystyle 1 + i$, which is in the first quadrant.
$\displaystyle r = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$\displaystyle \theta = \arctan{\frac{1}{1}} = \frac{\pi}{4}$.
Therefore $\displaystyle 1 + i = \sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right) = \sqrt{2}\,e^{\frac{\pi}{4}i}$.
So $\displaystyle (1 + i)^4 = (\sqrt{2}\,e^{\frac{\pi}{4}i})^4$
$\displaystyle = 4\,e^{\pi i}$
$\displaystyle = -4$.
Can you see that
$\displaystyle \log{[(1 + i)^4]} = \log{(-4)}$
$\displaystyle = \ln{|-4|} + i\arg{(-4)}$
$\displaystyle = \ln{4} + \pi i$
$\displaystyle = 2\ln{2} + \pi i$.
Now it is written in terms of its real and imaginary parts.