Question

If $i - 1$ then $- i$

• A. 1
• B. 2
• C. $n$
• D. $\left( \frac{i - 1}{i + 1} \right)^{n}$

Answer 1

Answer: (C)

$n$

Answer 2

Given that $y = - 1$, therefore

$x$

$= \frac { ( 1 + b ) ^ { 2 } - a ^ { 2 } + 2 i a ( 1 + b ) } { 1 + b ^ { 2 } + 2 b + a ^ { 2 } }$ $z_{1}$

$z_{2}$

Trick : Put $(z_{1}z_{2}) = {Re}(z_{1}){Re}(z_{2}) - {Im}(z_{1}){Im}(z_{2})$, $\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right)\left( \frac{3 + 4i}{2 - 4i} \right)$

But options (1) and (3) give 1.

So again put $= \left\lbrack \frac{1 + 2i}{1^{2} + 2^{2}} + \frac{3 - 3i}{1^{2} + 1^{2}} \right\rbrack\left\lbrack \frac{6 - 16 + 12i + 8i}{2^{2} + 4^{2}} \right\rbrack$.

Which gives (3) only.