# Find the modulus and argument of the complex numbers.

\[\dfrac{{5 - i}}{{2 - 3i}}\]

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**Hint:**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.