$\left(\frac{1+i}{1-i}\right)^{3}$ কে $\mathrm{A}+i \mathrm{~B}$ আাকারে প্রকাশ করলে নিচের কোনটি সঠিক ?

Step1.

$\begin{array}{l} \frac{1+i}{1-i} \cdot \frac{1+i}{1+i}=\frac{(1+i)^{2}}{(1-i)(1+i)} \\ (1-i)(1+i)=1^{2}-i^{2}=1-(-1)=1+1=2 \end{array}$

Now, $(1+i)^{2}=1^{2}+2 i+i^{2}=1+2 i-1=2 i$

So, the fraction simplifies, $\frac{2 i}{2}=i$.

step-02. $\quad i ^{3}=i+i+i=i^{2} \cdot i=(-1) \cdot i=-i$

step-03. Expross in the form $A+i B$

the result of $\left(\frac{1+i}{1-i}\right)^{3}$ simplifiers to $-i$ can be written as $0-i$ or $O+(-1)i$.

Thus $A=0$ and $B=-1$ the correct option $0+i(-1)$