Question

Express $\frac{(1+i{)}^{2n+1}}{(1-i{)}^{2n-1}},n\in N$ in the modulus-amplitude form

• A. $\{\begin{array}{c}2(cos{0}^o+isin{0}^o)\text{if n is even}\\ 2(cos\pi -isin\pi )\text{if n is odd}\end{array}$
• B. $\{\begin{array}{c}2(cos{0}^o-isin{0}^o)\text{if n is even}\\ 2(cos\pi +isin\pi )\text{if n is odd}\end{array}$
• C. $\{\begin{array}{c}2(cos{0}^o+isin{0}^o)\text{if n is even}\\ 2(cos\pi +isin\pi )\text{if n is odd}\end{array}$
• D. $\{\begin{array}{c}2(cos{0}^o-isin{0}^o)\text{if n is even}\\ 2(cos\pi -isin\pi )\text{if n is odd}\end{array}$

Answer 1

The correct option is C $\{\begin{array}{c}2(cos{0}^o+isin{0}^o)\text{if n is even}\\ 2(cos\pi +isin\pi )\text{if n is odd}\end{array}$

The equation can be simplified to:
${\left(\frac{1+i}{1-i}\right)}^{2n}.(1+i).(1-i)={\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)}^{2n}.2={\left(\frac{2i}{2}\right)}^{2n}.2=2.i^{2n}=2.{(-1)}^{n}neven:2=2(cos{0}^o+isin{0}^o)nodd:-2=2(cos\pi +isin\pi )$
Hence, (c) is correct.