Question

How do you divide $\dfrac{{1 + 7i}}{{9 - 5i}}$ in trigonometric form?

Answer 1

Hint : In order to solve the question given above, you need to have knowledge about complex numbers. It is defined as a number that can be expressed in the form $a + bi$ , where $a$ and $b$ are both real numbers whereas $i$ is an imaginary unit. Also, you need to know the properties and formulas used in trigonometry.

Formula used:
To solve the above question, you need to remember the trigonometric formulas:
$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ .
$\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ .

Complete step by step solution:
We have to divide $\dfrac{{1 + 7i}}{{9 - 5i}}$ in trigonometric form.
First, let us write the two complex numbers in polar coordinates. This gives us:
${z_1} = {r_1}\left( {\cos \alpha + i\sin \alpha } \right)$ and ${z_2} = {r_2}\left( {\cos \beta + i\sin \beta } \right)$ .
Let us take the two complex numbers as ${a_1} + {b_1}i$ and ${a_2} + {b_2}i$ .
Also, ${r_1} = \sqrt {{a_1}^2 + {b_1}^2}$ ; ${r_2} = \sqrt {{a_2}^2 + {b_2}^2}$ ; $\alpha = {\tan ^{ - 1}}\left( {\dfrac{{{b_1}}}{{{a_1}}}} \right)$ and $\beta = {\tan ^{ - 1}}\left( {\dfrac{{{b_2}}}{{{a_2}}}} \right)$ .
On dividing, we get:
$\left[ {\dfrac{{{r_1}}}{{{r_2}}}} \right] \left[ {\dfrac{{\cos \alpha + i\sin \alpha }}{{\cos \beta + i\sin \beta }}} \right]$ .
On solving,
$\left[ {\dfrac{{{r_1}}}{{{r_2}}}} \right] \left[ {\dfrac{{\cos \alpha + i\sin \alpha }}{{\cos \beta + i\sin \beta }} \times \dfrac{{\cos \beta - i\sin \beta }}{{\cos \beta - i\sin \beta }}} \right]$
$\Rightarrow \left[ {\dfrac{{{r_1}}}{{{r_2}}}} \right] \left[ {\dfrac{{\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right) + i\left( {\sin \alpha \cos \beta - \cos \alpha \sin \beta } \right)}}{{{{\cos }^2}\beta + {{\sin }^2}\beta }}} \right]$ .
Now solve the above brackets using the AB formulas of trigonometry where: $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ and $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ .
So, the above equation becomes:
$\left[ {\dfrac{{{r_1}}}{{{r_2}}}} \right] \left[ {\cos \left( {\alpha - \beta } \right) + i\sin \left( {\alpha - \beta } \right)} \right]$ .
Now, $\dfrac{{{z_1}}}{{{z_2}}}$ is given by $\left( {\dfrac{{{r_1}}}{{{r_2}}},\left( {\alpha - \beta } \right)} \right)$ . Therefore, to divide complex numbers by ${z_1}$ and ${z_2}$ .
Now, take new angle as $\left( {\alpha - \beta } \right)$ and ratio $\dfrac{{{r_1}}}{{{r_2}}}$ of the modulus of two numbers.
In the above question, $1 + 7i$ can be written as ${r_1}\left( {\cos \alpha + i\sin \alpha } \right)$ where
${r_1} = \sqrt {{1^2} + {7^2}}$
$= \sqrt {50}$ .
And $\alpha = {\tan ^{ - 1}}7$ .
And for $9 - 5i$ can be written as ${r_2}\left( {\cos \beta + i\sin \beta } \right)$ where
${r_2} = \sqrt {{9^2} + {5^2}}$
$= \sqrt {106}$ .
And $\beta = {\tan ^{ - 1}}\left( {\dfrac{{ - 5}}{9}} \right)$ .
Now, $\dfrac{{{z_1}}}{{{z_2}}} = \dfrac{{\sqrt {50} }}{{\sqrt {106} }}\left( {\cos \left( {\alpha - \beta } \right) + i\sin \left( {\alpha - \beta } \right)} \right)$ .
So,
$\tan \left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}$
$= \dfrac{{7 - \left( {\dfrac{{ - 5}}{9}} \right)}}{{1 + 7\left( {\dfrac{{ - 5}}{9}} \right)}}$
$= \dfrac{{\dfrac{{68}}{9}}}{{\dfrac{{ - 26}}{9}}}$ .
Now, we get:
$\tan \left( {\alpha - \beta } \right) = - \dfrac{{34}}{{13}}$ .
Hence, $\dfrac{{1 + 7i}}{{9 - 5i}} = \dfrac{{\sqrt {50} }}{{\sqrt {106} }}\left( {\cos \theta + i\sin \theta } \right)$ , where $\theta = {\tan ^{ - 1}}\left( { - \dfrac{{34}}{{13}}} \right)$ .

Note : While solving questions similar to the one given above, you need to remember the concept of complex numbers. You need to remember the formulas used in trigonometry. For example: in the above question we have used $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ and $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ .