Question

Evaluate:

${\left(\begin{array}{c}\frac{1+\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}}{1+\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}}\end{array}\right)}^4$

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

Answer 1

Answer: (B)

$-1$

Let, $1+cos\frac{\pi }{4}=2co{s}^2\frac{\pi }{8}$ and ..........(1)

$sin\frac{\pi }{4}=2sin\frac{\pi }{8}cos\frac{\pi }{8}$.......(2)

GIven: ${\left(\frac{1+cos\frac{\Pi }{4}-isin\frac{\Pi }{4}}{1+cos\frac{\Pi }{4}+isin\frac{\Pi }{4}}\right)}^4$

$={\left(\frac{2co{s}^2\left(\pi /8\right)-i2sin\left(\pi /8\right)cos\left(\pi /8\right)}{2co{s}^2\left(\pi /8\right)+i2sin\left(\pi /8\right)cos\left(\pi /8\right)}\right)}^4$

$={\left(\frac{2cos\left(\pi /8\right)\left[cos\right(\pi /8)-isin(\pi /8\left)\right]}{2cos\left(\pi /8\right)\left[cos\right(\pi /8)+isin(\pi /8\left)\right]}\right)}^4$

$={\left(\frac{e^{-i\left(\pi /8\right)}}{e^{i\left(\pi /8\right)}}\right)}^4=(e^{-i\left(\pi /4\right)}{)}^4=e^{-i\pi }=cos\pi -isin\pi =-1.$