Question

The value of ${\left(\frac{1+cos\frac{\pi }{2}-isin\frac{\pi }{2}}{1+cos\frac{\pi }{2}+isin\frac{\pi }{2}}\right)}^2$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $-1$

Let,

$1+cos\frac{\pi }{2}=2co{s}^2\frac{\pi }{4}$ ..........(1)

and

$sin\frac{\pi }{2}=2sin\frac{\pi }{4}cos\frac{\pi }{4}$.......(2)

GIven:

${\left(\frac{1+cos\frac{\Pi }{2}-isin\frac{\Pi }{2}}{1+cos\frac{\Pi }{2}+isin\frac{\Pi }{2}}\right)}^2$

$={\left(\frac{2co{s}^2\left(\pi /4\right)-i2sin\left(\pi /4\right)cos\left(\pi /4\right)}{2co{s}^2\left(\pi /4\right)+i2sin\left(\pi /4\right)cos\left(\pi /4\right)}\right)}^2$

$={\left(\frac{2cos\left(\pi /4\right)\left[cos\right(\pi /4)-isin(\pi /4\left)\right]}{2cos\left(\pi /4\right)\left[cos\right(\pi /4)+isin(\pi /4\left)\right]}\right)}^2$

$={\left(\frac{e^{-i\left(\pi /4\right)}}{e^{i\left(\pi /4\right)}}\right)}^2$

$=(e^{-i\left(\pi /2\right)}{)}^2=e^{-i\pi }$

$=cos\pi -isin\pi$

$=-1.$