Question

Let $z=\frac{11-3i}{1+i}.$ If a is a real number such that $z-i\alpha$ is real, then the value of $\alpha$ is

• A. 4
• B. $-\,4$
• C. 7
• D. $-\text{ }7$

Answer 1

Answer: (D)

$-\text{ }7$

Answer 2

$\because$ $z=\frac{11-3i}{1+i}\times \frac{1-i}{1-i}$
$=\frac{11-11i-3i-3}{1+1}$
$=\frac{8-14i}{2}=4-7i$
Also, $\alpha$ is a real number such that $z-i\alpha$ is real.
$\therefore$ $4-7i-i\alpha$ is real, if $\alpha =-7$ .