Question

The nonzero real value of $x$ for which $\frac{(1+ix)(1+2ix)}{1-ix}$ is purely real is

• A. $\sqrt{2}$
• B. $1$
• C. $-\sqrt{2}$
• D. none of these

Answer 1

Answer: (A), (C)

$-\sqrt{2}$
$\sqrt{2}$

The value of $\frac{(1+ix)(1+2ix)}{1-ix}$

$=\frac{(1+ix{)}^2(1+2ix)}{1+x^2}$

$=\frac{(1-x^2+2ix)(1+2ix)}{1+x^2}$

$=\frac{(1+2ix{)}^2-x^2(1+2ix)}{1+x^2}$

$=\frac{1-4x^2+4ix-x^2-2i{x}^3}{1+x^2}$

$=\frac{1-5x^2+2ix(2-x^2)}{1+x^2}$

Hence the above complex number is purely real if

$2x(2-x^2)=0$

$x=0$ and $x^2-2=0$

Thus $x=-\sqrt{2},0,\sqrt{2}$.