Question

If ${\left(\frac{1+i}{1-i}\right)}^3-{\left(\frac{1-i}{1+i}\right)}^3=A+iB$, then $A,B=$

• A. $0,2$
• B. $0,-2$
• C. $2,0$
• D. $-2,0$

Answer 1

Answer: (B)

$0,-2$

$\frac{1+i}{1-i}=\frac{(1+i{)}^2}{1^2-i^2}=i$

$\frac{1-i}{1+i}=\frac{(1-i{)}^2}{1^2-i^2}=-i$

Putting the above values obtained in the given equation

$i^3-(-i{)}^3(=A+iB)$

$=-i-(+i)$

$=0-2i$

$\therefore A=0,B=-2$