Question

Express the complex number $\dfrac{{(1 + 7i)}}{{{{(2 - i)}^2}}}$ in the polar form.

Answer 1

Hint: Rationalize the number first to remove radicals and then simplify and compare with polar form and substitute values.
So, first we remove the radical in the denominator such that the calculation becomes easy. To do this we rationalize the number and remove the radical. Then, after that we simplify and compare it with the polar form and obtain the required values. We use the property which say ${i^2} = - 1$ Then after obtaining them, substitute in the polar form. Thus, we get to represent the complex number in polar form

Complete step by step solution:
First, we begin by rationalizing such that the denominator is rid of radical value.
We multiply ${(2 + i)^2}$ on both the numerator and denominator
\[
  \dfrac{{(1 + 7i)}}{{{{(2 - i)}^2}}} = \dfrac{{(1 + 7i)}}{{{{(2 - i)}^2}}} \times \dfrac{{{{(2 + i)}^2}}}{{{{(2 + i)}^2}}} \\
   = \dfrac{{(1 + 7i){{(2 + i)}^2}}}{{{{(2 - i)}^2}{{(2 + i)}^2}}} \\
   = \dfrac{{(1 + 7i)(4 - i + 4i)}}{{{{((2 - i)(2 + i))}^2}}} \\
   = \dfrac{{(4 - 1 + 4i + 28i - 7i - 28)}}{{{{((2 - i)(2 + i))}^2}}} \\
 \]
For the denominator, there is an identity which can used to simplify it which is
\[\;\;{\text{ }}\;\;[(a + ib)(a - ib) = {a^2} + {b^2}]\]
This identity is applied in the denominator
$
   = \dfrac{{( - 25 + 25i)}}{{{{({2^2} + {1^2})}^2}}} \\
   = - 1 + i \\
$
If θ is principal argument and r is magnitude of complex number z then Polar form is represented by:
\[z = r(cos\theta + isin\theta )\]

Now, we compare this form, with the obtained complex number form and get the value of r and $\theta $.
After comparison of both equations, we get

\[
   - 1 = rcos\theta \;{\text{ }} \\
  \;1 = rsin\theta \\
\]
From the above two equations, we can calculate the value of r.
This can be done by squaring and adding on both sides. Then we get
\[{r^2}(co{s^2}\theta + si{n^2}\theta ) = {( - 1)^2} + {1^2} = 2\]
Now further, we get the value of r, which is
\[\;r = \sqrt 2 \]

Now, we are going to find the value of $\theta $. This can be done by,
$\dfrac{{r\sin \theta }}{{r\cos \theta }} = \dfrac{1}{{ - 1}}$

The r in the numerator and denominator gets cancelled and the LHS gets converted to $\tan \theta $.
$
  \tan \theta = - 1 \\
  \tan \theta = - \tan \left( {\dfrac{\pi }{4}} \right) \\
  \theta = \dfrac{{3\pi }}{4}\because [\tan (\pi - \theta ) = - \tan \theta ] \\
$
Now, we have both r and $\theta $ value. Now, we substitute them into the polar form equation and get the final equation which is the answer.
$z = \sqrt 2 \left( {\cos \left( {\dfrac{{3\pi }}{4}} \right) + i\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)$
The above is the polar form of the given complex number.

Note: We need to know the necessary values of some predefined values and for the sum to start it is necessary to rationalize the complex number such that the radicals in the denominator get eliminated from the number.