Question

If $i^{i^{-\ast }}=A+iB(i=iota=\sqrt{-1})$ then

• A. $\tan \frac{\pi }{2}A=\frac{B}{A}$
• B. $\tan \frac{\pi }{2}A=\frac{A}{B}$
• C. $A^2+B^2=e^{-\pi B}$
• D. $A^2+B^2=e^{-\pi A}$

Answer 1

Answer: (A), (C)

Correct answer is (a) $\tan \frac{\pi }{2}A=\frac{B}{A}$ Correct answer is (c) $A^2+B^2=e^{-\pi B}$