Question

Convert the following in the polar form:
(a) $\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}$
(b) $\dfrac{{1 + 3i}}{{1 - 2i}}$

Answer 1

In order to answer this question, we have to convert both the given expressions involving complex numbers into their polar form. We should first remove the complex numbers from the denominators of the given expression and then work them to convert them into polar form. Polar form of a complex number is of the form $\left( {r\cos \theta } \right) + i\left( {r\sin \theta } \right)$.

Complete step by step answer:
The given expressions involving complex numbers are in rectangular form. So, we are asked to convert these complex numbers into polar form. A complex number $a + ib$ is converted into polar form as $\left( {r\cos \theta } \right) + i\left( {r\sin \theta } \right)$, where r is the modulus of a complex number and $\theta$ is the argument of the complex number.

(a) Now, let the given complex number be Z.
So, we have, $Z = \dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}$
We will first expand the whole square in the denominator using the algebraic identity ${\left( {a - b} \right)^2} = \left( {{a^2} - 2ab + {b^2}} \right)$ . So, we get,
$\Rightarrow Z = \dfrac{{1 + 7i}}{{{{\left( 2 \right)}^2} + {i^2} - 2\left( 2 \right)\left( i \right)}}$
We know that the value of ${i^2}$ is $\left( { - 1} \right)$. So, we get,
$\Rightarrow Z = \dfrac{{1 + 7i}}{{4 - 1 - 2\left( 2 \right)\left( i \right)}}$
$\Rightarrow Z = \dfrac{{1 + 7i}}{{3 - 4i}}$

Now, to remove complex numbers from the denominator, we multiply and divide the expression by $3 + 4i$. So, we get,
$\Rightarrow Z = \dfrac{{1 + 7i}}{{3 - 4i}} \times \dfrac{{3 + 4i}}{{3 + 4i}}$
$\Rightarrow Z = \dfrac{{3 + 21i + 4i + 28{i^2}}}{{{3^2} - {{\left( {4i} \right)}^2}}}$
We know that the value of ${i^2}$ is $\left( { - 1} \right)$. So, we get,
$\Rightarrow Z = \dfrac{{3 + 25i - 28}}{{9 - \left( { - 16} \right)}}$
Simplifying the expression, we get,
$\Rightarrow Z = \dfrac{{ - 25 + 25i}}{{25}}$
$\Rightarrow Z = - 1 + i$
So, the complex number $\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}$ can be simplified as $\left( { - 1 + i} \right)$.

Now, we convert the complex number to polar form.
We find the modulus of the complex number $\left( { - 1 + i} \right)$.
The modulus of a complex number is given by $\left| Z \right|$ and it is calculated as:
$\left| Z \right| = \sqrt {{x^2} + {y^2}}$
Thus, putting in the values of x and y, we get the modulus of given complex number as:
$\Rightarrow \left| Z \right| = \sqrt {{{( - 1)}^2} + {{(1)}^2}}$
$\Rightarrow \left| Z \right| = \sqrt 2$
So, the modulus of a given complex number is $\sqrt 2$.
Now, we find the argument of the complex number.
So, $- 1 + i = \sqrt 2 \left( {\cos \theta + i\sin \theta } \right)$
Comparing both sides of equation, we get the values of cosine and sine,
$\cos \theta = \dfrac{{ - 1}}{{\sqrt 2 }}$ and $\sin \theta = \dfrac{1}{{\sqrt 2 }}$.
Now, we know that sine is positive and cosine is negative in the second quadrant.
Also, $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
So, $\theta = \pi - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4}$.

Therefore, the polar form of the complex number $\dfrac{{1 + 7i}}{{{{\left( {2 - i} \right)}^2}}}$ is $\sqrt 2 \left( {\cos \left( {\dfrac{{3\pi }}{4}} \right) + i\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)$.

(b) Let the given complex number be $P$.
So, we have, $P = \dfrac{{1 + 3i}}{{1 - 2i}}$
Now, to remove complex numbers from the denominator, we multiply and divide the expression by $1 + 2i$. So, we get,
$\Rightarrow P = \dfrac{{1 + 3i}}{{1 - 2i}} \times \dfrac{{1 + 2i}}{{1 + 2i}}$
We know that the value of ${i^2}$ is $\left( { - 1} \right)$. So, we get,
$\Rightarrow P = \dfrac{{1 + 2i + 3i + 6{i^2}}}{{1 - \left( {4{i^2}} \right)}}$
$\Rightarrow P = \dfrac{{1 + 5i - 6}}{{1 + 4}}$
Simplifying the expression, we get,
$\Rightarrow P = \dfrac{{ - 5 + 5i}}{5}$
$\Rightarrow P = - 1 + i$
So, the complex number $P = \dfrac{{1 + 3i}}{{1 - 2i}}$ can be simplified as $P = - 1 + i$.

Now, we convert the complex number to polar form.
We find the modulus of the complex number $P = - 1 + i$.
The modulus of a complex number is given by $\left| Z \right|$ and it is calculated as:
$\left| Z \right| = \sqrt {{x^2} + {y^2}}$
Thus, putting in the values of x and y, we get the modulus of given complex number as:
$\Rightarrow \left| P \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}}$
$\Rightarrow \left| P \right| = \sqrt 2$
So, the modulus of a given complex number is $\sqrt 2$.
Now, we find the argument of the complex number.
So, $- 1 + i = \sqrt 2 \left( {\cos \theta + i\sin \theta } \right)$
Comparing both sides of equation, we get the values of cosine and sine,
$\cos \theta = \dfrac{{ - 1}}{{\sqrt 2 }}$ and $\sin \theta = \dfrac{1}{{\sqrt 2 }}$.
Now, we know that sine is positive and cosine is negative in the second quadrant.
Also, $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
So, $\theta = \pi - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4}$.

Therefore, the polar form of the complex number $\dfrac{{1 + 3i}}{{1 - 2i}}$ is $\sqrt 2 \left( {\cos \left( {\dfrac{{3\pi }}{4}} \right) + i\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)$.

Note: Complex numbers can also be interpreted in polar form, which associates a complex number with its distance from the origin (its magnitude) and a specific angle called the complex number's argument. Complex numbers are geometrically associated with the complex plane, which is an Argand plane. We must remember the range of arguments of a complex number as $( - \pi ,\pi ]$ to tackle such problems.