Question

How do you divide $\dfrac{7-i}{3-i}$ in trigonometric form?

Answer 1

Hint: We are given 2 complex numbers, 7-i and 3-i and are asked to divide them in the trigonometric form. To do so, we will first learn what trigonometric form of complex numbers is. We will learn how to find the argument and magnitude of it and we will learn how division will be carried out in such forms. We will start solving by considering ${{z}_{1}}=7-i$ and ${{z}_{2}}=3-i$

Complete step-by-step answer:
We are given two complex numbers, 7-i and 3-i and are asked to divide them in the trigonometric form. Before, we proceed, let us first learn about the trigonometric form of complex numbers. For any complex number z= a+ ib, the trigonometric form is given as $z=r\left( \cos \theta +i\sin \theta \right)$. Where r is called the magnitude and $\theta $ is called the argument.
Let us consider ${{z}_{1}}=7-i$ and ${{z}_{2}}=3-i$.
So, we will now convert them into its trigonometric forms.
We know that for any complex number, z= x+ iy, its magnitude, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and the argument, $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$.
So, for ${{z}_{1}}=7-i$,
$r=\sqrt{{{7}^{2}}+\left( -{{i}^{2}} \right)}=\sqrt{49+1}=\sqrt{50}=7.07$
 $\theta ={{\tan }^{-1}}\left( \dfrac{-{{i}^{2}}}{7} \right)=-{{\tan }^{-1}}\left( \dfrac{1}{7} \right)$, lies in the 4th quadrant.
$\theta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)=0.142$, lies in the 4th quadrant.
So, our actual $\theta =2\pi -\alpha =2\pi -0.142=6.14$.
Hence, we get r = 7.07 and $\theta =6.14$.
So, ${{z}_{1}}=7.07\left[ \cos \left( 6.14 \right)+i\sin \left( 6.14 \right) \right]$ in the trigonometric form.
Similarly, we will now consider the second complex number. We have, ${{z}_{2}}=3-i$.
So, we get,
$r=\sqrt{{{3}^{2}}+\left( -{{i}^{2}} \right)}=\sqrt{9+1}=\sqrt{10}=3.16$
 $\theta ={{\tan }^{-1}}\left( \dfrac{-{{i}^{2}}}{3} \right)=-{{\tan }^{-1}}\left( \dfrac{1}{3} \right)$, lies in the 4th quadrant.
$\theta ={{\tan }^{-1}}\left( \dfrac{1}{3} \right)=0.322$, lies in the 4th quadrant.
So, our actual $\theta =2\pi -\alpha =2\pi -0.322=5.96$.
Hence, we get r = 3.16 and $\theta =5.96$.
Hence, ${{z}_{2}}=3.16\left[ \cos \left( 5.96 \right)+i\sin \left( 5.96 \right) \right]$ in the trigonometric form.
Now, we have to find the division of these two terms, that is for \[{{z}_{1}}=r\left[ \cos \left( a \right)+i\sin \left( a \right) \right]\] and \[{{z}_{2}}=w\left[ \cos \left( b \right)+i\sin \left( b \right) \right]\], we have to find $\dfrac{{{z}_{1}}}{{{z}_{2}}}$, that is $=\dfrac{r}{w}\left[ \cos \left( a-b \right)+i\sin \left( a-b \right) \right]$.
Therefore, considering our question, we get,
 $\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{7.07\left[ \cos \left( 6.14 \right)+i\sin \left( 6.14 \right) \right]}{3.16\left[ \cos \left( 5.96 \right)+i\sin \left( 5.96 \right) \right]}$
We can write it as,
$\dfrac{{{z}_{1}}}{{{z}_{2}}}=\dfrac{7.07}{3.16}\left[ \cos \left( 6.14-5.96 \right)+i\sin \left( 6.14-5.96 \right) \right]$
On simplifying, we get,
$\dfrac{{{z}_{1}}}{{{z}_{2}}}=2.236\left[ \cos \left( 0.18 \right)+i\sin \left( 0.18 \right) \right]$
If we change it back to complex form, we can write,
$\dfrac{{{z}_{1}}}{{{z}_{2}}}=2.2+0.4i$

Note: We need to be careful with the signs. According to the sign, the quadrant of the argument is decided. If sin and cos are both positive, then it lies in the ${{1}^{st}}$ quadrant. If sin is positive and cos is negative, it lies in the ${{2}^{nd}}$ quadrant. If sin and cos are negative it lies in the ${{3}^{rd}}$ quadrant. If sin is negative and cos is positive, it lies in the ${{4}^{th}}$ quadrant. An argument of the ${{4}^{th}}$ quadrant is given as $2\pi -\theta $.