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 \[\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b\] .