Question

The value of ${log}_{1-i}(1+i)$ is

• A. $\frac{log\sqrt{2}+i\frac{\pi }{4}}{log\sqrt{2}-i\frac{\pi }{4}}$
• B. $\frac{log\sqrt{2}-i\frac{\pi }{4}}{log\sqrt{2}+i\frac{\pi }{4}}$
• C. $\frac{log2+i\frac{\pi }{4}}{log2-i\frac{\pi }{4}}$
• D. None of these

Answer 1

The correct option is A $\frac{log\sqrt{2}+i\frac{\pi }{4}}{log\sqrt{2}-i\frac{\pi }{4}}$

Given:

${log}_{1-i}(1+i)=\frac{log(1-i)}{log(1+i)}=\frac{log\sqrt{2}{e}^{-i\left(\pi /4\right)}}{log\sqrt{2}{e}^{i\left(\pi /4\right)}}=\frac{log\sqrt{2}+log\left(e^{-i\left(\pi /4\right)}\right)}{log\sqrt{2}+log\left(e^{i\left(\pi /4\right)}\right)}=\frac{log\sqrt{2}-i\frac{\pi }{4}}{log\sqrt{2}+i\frac{\pi }{4}}$.