Question

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

• A. $4$
• B. $-4$
• C. $7$
• D. $-7$

Answer 1

The correct option is D

$-7$

Explanation for the correct option:

Calculating the value of $a$:

Given,

$z=\frac{(11-3i)}{(1+i)}$ and $z-ia$ is real.

Simplifying $z$:

$\begin{array}{rcl}z& =& \frac{11-3i}{1+i}\\ & =& \frac{11-3i}{1+i}\times \frac{1-i}{1-i}\\ & =& \frac{(11-3i)(1-i)}{1-i^2}\\ & =& \frac{-14i+8}{2}\mathbf{[}\mathbf{\because }{\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{]}\\ & =& 4-7i\end{array}$

Now,

$$\begin{gathered} z-ia=4-7i-ia \\ =4-i(7+a) \end{gathered}$$

Given, $z-ia$ is real. So, the imaginary part of $z-ia$ will be zero.

Hence,

$$\begin{gathered} 7+a=0 \\ \Rightarrow a=-7 \end{gathered}$$

Therefore, the correct answer is option (D).