Question

express (1-i^3/1+i^3) ^2 in the form of a+ib. where a, b€R. state the value of a and b​

Answer 1

Answer:

The answer is $a=-1, b=0$

Step-by-step explanation:

$(\frac{1-i^3}{1+i^3})^2=(\frac{1-i(i^2)}{1+i(i^2)})^2\\=(\frac{1-i(-1)}{1+i(-1)})^2=(\frac{1+i}{1-i})^2$

(since $i=\sqrt{-1}\Rightarrow i^2=-1$)

$=\frac{(1+i)^2}{(1-i)^2}=\frac{1+i^2+2i}{1+i^2-2i}\\=\frac{1-1+2i}{1-1-2i}=\frac{2i}{-2i}=-1\\=-1+0i$

Thus $a=-1, b=0$