Question

Find the principal and general value of ${\log }_e(-1+i)$

• A. $\frac{1}{2}\log 2+\frac{3}{4}\pi i,\frac{1}{2}{\log }_{e}2+(n+\frac{3}{4})\pi i$ ($n$ integer)
• B. $\frac{1}{2}\log 2+\frac{3}{4}\pi i,\frac{1}{2}{\log }_{e}2+(2n+\frac{3}{4})\pi i$ ($n$ integer)
• C. $\frac{1}{2}\log 2+\frac{1}{4}\pi i,\frac{1}{2}{\log }_{e}2+(2n+\frac{1}{4})\pi i$ ($n$ integer)
• D. $\frac{1}{2}\log 2+\frac{1}{4}\pi i,\frac{1}{2}{\log }_{e}2+(n+\frac{1}{4})\pi i$ ($n$ integer)

Answer 1

Answer: (B)

$\frac{1}{2}\log 2+\frac{3}{4}\pi i,\frac{1}{2}{\log }_{e}2+(2n+\frac{3}{4})\pi i$ ($n$ integer)

Given ${\log }_e(-1+i)$

$={\log }_e|-1+i|+iArg(-1+i)$

$={\log }_e\sqrt{2}+i\frac{3\pi }{4}\Rightarrow \frac{1}{2}{\log }_{e}2+i\frac{3\pi }{4}$

This is the principal solution.

General Solution is

$\frac{1}{2}{\log }_{e}2+i(2n\pi +\frac{3\pi }{4})$ ($n$ is integer)