Question

Find the modulus and argument of the complex number $z = \dfrac{{1 - 2i}}{{1 + 3i}}$ in polar form.

Answer 1

Hint: First, multiply and divide by the conjugate of the denominator to remove the imaginary part from the denominator and simplify it. After that find the modulus of z by the formula $\left| z \right| = \sqrt {{a^2} + {b^2}} $. Then, find the argument of z by ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$.

Complete step by step answer:
Given, $z = \dfrac{{1 - 2i}}{{1 + 3i}}$
Multiply and divide z with the conjugate of the denominator,
$\Rightarrow z = \dfrac{{1 - 2i}}{{1 + 3i}} \times \dfrac{{1 - 3i}}{{1 - 3i}}$
Multiply the terms on the right to get real numbers in the denominator,
$\Rightarrow z = \dfrac{{1 + 6{i^2} - 2i - 3i}}{{1 - 9{i^2}}}$
As we know ${i^2} = - 1$, substitute the value in the equation,
$\Rightarrow z = \dfrac{{1 + 6\left( { - 1} \right) - 2i - 3i}}{{1 - 9\left( { - 1} \right)}}$
Open the brackets and add the like terms,
$\Rightarrow z = \dfrac{{ - 5 - 5i}}{{10}}$
Separate the real part and imaginary parts,
$\Rightarrow z = - \dfrac{5}{{10}} - i\dfrac{5}{{10}}$
Cancel out the common factors,
$\Rightarrow z = - \dfrac{1}{2} - \dfrac{1}{2}i$
The formula of modulus is,
$\Rightarrow \left| z \right| = \sqrt {{a^2} + {b^2}} $
Here $a = - \dfrac{1}{2}$ and $b = - \dfrac{1}{2}$. Then,
$\Rightarrow \left| z \right| = \sqrt {{{\left( { - \dfrac{1}{2}} \right)}^2} + {{\left( { - \dfrac{1}{2}} \right)}^2}} $
Square the terms in the bracket,
$\Rightarrow \left| z \right| = \sqrt {\dfrac{1}{4} + \dfrac{1}{4}} $
Since the denominator is the same. So, add the numerator,
$\Rightarrow \left| z \right| = \sqrt {\dfrac{2}{4}} $
Cancel out the common factors from the numerator and denominator,
$\Rightarrow \left| z \right| = \dfrac{1}{{\sqrt 2 }}$
Now, $\tan \alpha = \left| {\dfrac{b}{a}} \right|$. Then,
$\Rightarrow \tan \alpha = \left| {\dfrac{{ - \dfrac{1}{2}}}{{ - \dfrac{1}{2}}}} \right|$
Cancel out the common factor,
$\Rightarrow \tan \alpha = 1$
Take ${\tan ^{ - 1}}$ on both sides,
$
  \alpha = {\tan ^{ - 1}}1 \\
   = \dfrac{\pi }{4} \\
$
As both of the real part and the imaginary part of the complex number is negative. The number will lie in 3rd quadrant. Then,
$\Rightarrow \arg \left( z \right) = \pi + \alpha $
Substitute the value of $\alpha $,
$\Rightarrow \arg \left( z \right) = \pi + \dfrac{\pi }{4}$
Add the terms on the right side,
$\Rightarrow \arg \left( z \right) = \dfrac{{5\pi }}{4}$

Hence, the modulus is $\dfrac{1}{{\sqrt 2 }}$ and the argument is $\dfrac{{5\pi }}{4}$.

Note:
The complex numbers are the field C of numbers of the form $x + iy$, where x and y are real numbers and i is the imaginary unit equal to the square root of -1. When a single letter z is used to denote a complex number. It is denoted as $z = x + iy$.