Question

The complex number $\frac{\left(-\sqrt{3}+3i\right)\left(i-1\right)}{\left(3+\sqrt{3i}\right)\left(i\right)\left(\sqrt{3}+\sqrt{3i}\right)}$when represented in Argand diagram lies in which of the following.

• A. In the first quadrant
• B. In the second quadrant
• C. On the Y-axis (imaginary axis)
• D. On the X- axis (real axis)

Answer 1

The correct option is C

On the Y-axis (imaginary axis)

Explanation of correct answer :

Determining where complex number $z$ lies by simplifying the expression:

Given, $z=$$\frac{\left(-\sqrt{3}+3i\right)\left(i-1\right)}{\left(3+\sqrt{3i}\right)\left(i\right)\left(\sqrt{3}+\sqrt{3i}\right)}$

Taking $\sqrt{3}$ outside,

$\begin{array}{rcl}z& =& \frac{\sqrt{3}\left(-1+\sqrt{3i}\right)\left(i-1\right)}{{\left(\sqrt{3}\right)}^2\left(\sqrt{3}+i\right)\left(i\right)\left(1+i\right)}\\ & =& \frac{1}{\sqrt{3}}\frac{\left(-1+\sqrt{3i}\right)(i-1)}{\left(-1+\sqrt{3}i\right)(i+1)}\\ & =& \frac{1}{\sqrt{3}}\frac{(i-1)}{(i+1)}\end{array}$

Rationalize $z$, by multiplying its numerator and denominator by $(i-1)$.

$\begin{array}{rcl}z& =& \frac{1}{\sqrt{3}}\frac{(i-1)}{(i+1)}\frac{(i-1)}{(i-1)}\\ & =& \frac{1}{\sqrt{3}}\frac{i^2+1-2i}{i^2-1}\\ & =& -\frac{1}{\sqrt{3}}i(i^2=-1)\end{array}$

Hence, $z$ lies on Y-axis because it is purely imaginary.

Thus, the correct answer is Option (C).