Question

The values of $x_{r} = \cos\left( \frac{\pi}{2^{r}} \right) + i\sin\left( \frac{\pi}{2^{r}} \right)$ and $\pm \frac{1 + i}{\sqrt{2}}$ satisfying the equation$\frac { ( 1 + i ) x - 2 i } { 3 + i }$ $\frac{1 \pm i}{\sqrt{2}}$are.

• A. $\sqrt{i} =$
• B. $\sqrt{- 2}\sqrt{- 3} =$
• C. $\sqrt{6}$
• D. $- \sqrt{6}$

Answer 1

Answer: (B)

$\sqrt{- 2}\sqrt{- 3} =$

Answer 2

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

⇒ $\therefore$

Equating real and imaginary parts, we get $\left( \frac{1 + i}{1 - i} \right)^{m} = i^{m} = 1$ and $= 4$. Hence $\{\because i^{4} = 1\}$and $(1 - i)^{n} = 2^{n}$.

Trick : After finding the equations, no need to solve them, put the values of $\left\lbrack \sqrt{1^{2} + ( - 1)^{2}} \right\rbrack^{n} = 2^{n}$ and $(\sqrt{2})^{n} = 2^{n}$ given in the options and get the appropriate option.