Question

Find the modulus and argument of the complex numbers.
$\dfrac{{5 - i}}{{2 - 3i}}$

Answer 1

First we try to multiply the numerator and denominator with the conjugate of the denominator. Thus we reach such a term that the denominator becomes an integer. Then comparing with $r\cos \theta + ir\sin \theta$ we get the modulus and argument.

Complete step by step answer:

Consider the given complex number, $\dfrac{{5 - i}}{{2 - 3i}}$
By rationalization of given numbers.
$\dfrac{{5 - i}}{{2 - 3i}}$
Multiplying the numerator and denominator with the conjugate term of the denominator,
$= \dfrac{{(5 - i) \times (2 + 3i)}}{{(2 - 3i) \times (2 + 3i)}}$
On Simplifying, we get,
$= \dfrac{{10 - 2i + 15i - 3{i^2}}}{{{2^2} - {{(3i)}^2}}}$
Using ${{\text{i}}^{\text{2}}}{\text{ = ( - 1)}}$, we get,
$= \dfrac{{10 + 13i + 9}}{{4 + 9}}$
On simplifying we get,
$= \dfrac{{13 + 13i}}{{13}}$
On cancelling common terms we get,
$= 1 + i$

We have,
$\dfrac{{5 - i}}{{2 - 3i}} = 1 + i$

Let, $z = 1 + i$ which is of the form, $x + iy$ and here, $x = 1$and $y = 1$
Modulus of z $= \left| z \right|$ $= \sqrt {{x^2} + {y^2}}$
$= \sqrt {{1^2} + {1^2}}$
$= \sqrt 2$
Now, to find the argument, we take, $1 + i = r\cos \theta + ir\sin \theta$
So, we get by comparing, $1 = r\cos \theta$ and $1 = r\sin \theta$ where r is the modulus.
So, we have, $r = \sqrt 2$
Then, $\sin \theta = \cos \theta = \dfrac{1}{{\sqrt 2 }}$
So, now, we have both x and y positive, then, $\theta$ lies in the 1st quadrant.
$so,\theta = 45^\circ$
As $\sin {45^o} = \cos {45^o} = \dfrac{1}{{\sqrt 2 }}$
Hence, the argument of $z = \dfrac{\pi }{4}$.

Note: We can also solve the problem with the help of polar coordinates totally. The modulus can be found by comparing real and imaginary parts from the equation, $1 + i = r\cos \theta + ir\sin \theta$. Differentiating the real and imaginary and then by comparing them we can find the value of r, the value of r would give us our modulus, and then we can also find $\theta$ in the same process.