Question

How do you divide $\dfrac{4i+4}{6i+5}$ in trigonometric form?

Answer 1

Hint: To solve the given question first we will multiply the numerator and denominator of the given expression by complex conjugate of the denominator. Then we will find the modulus and the argument of obtained complex number. Then we will express the obtained complex number in polar form. The polar form or trigonometric form of complex number is given as
$z=x+iy=r\left( \cos \theta +i\sin \theta \right)$
Where, $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$

Complete step-by-step answer:
We have been given an expression $\dfrac{4i+4}{6i+5}$.
We have to divide the given expression in trigonometric form.
First let us multiply the numerator and denominator of given expression by complex conjugate of denominator. Then we will get
$\Rightarrow \dfrac{4i+4}{6i+5}\times \dfrac{6i-5}{6i-5}$
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{\left( 4i+4 \right)\left( 6i-5 \right)}{{{\left( 6i \right)}^{2}}-{{5}^{2}}} \\
 & \Rightarrow \dfrac{24{{i}^{2}}-20+24i-20i}{36{{i}^{2}}-25} \\
\end{align}\]
Now, we know that ${{i}^{2}}=-1$
Now, substituting the values we will get
\[\Rightarrow \dfrac{24\left( -1 \right)-20+24i-20i}{36\left( -1 \right)-25}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow \dfrac{-24-20+24i-20i}{-36-25} \\
 & \Rightarrow \dfrac{-44+4i}{-61} \\
\end{align}\]
Now, we can write the above obtained equation as
$\begin{align}
  & \Rightarrow \dfrac{-44}{-61}+\dfrac{4i}{-61} \\
 & \Rightarrow \dfrac{44}{61}-\dfrac{4i}{61} \\
\end{align}$
Now, we know that the modulus of complex number is given by
$\left| z \right|=\sqrt{{{\left( \dfrac{44}{61} \right)}^{2}}+{{\left( \dfrac{4}{61} \right)}^{2}}}$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow \left| z \right|=\sqrt{\dfrac{1936}{{{\left( 61 \right)}^{2}}}+\dfrac{16}{{{\left( 61 \right)}^{2}}}} \\
 & \Rightarrow \left| z \right|=\sqrt{\dfrac{1936+16}{{{\left( 61 \right)}^{2}}}} \\
 & \Rightarrow \left| z \right|=\sqrt{\dfrac{1952}{{{\left( 61 \right)}^{2}}}} \\
 & \Rightarrow \left| z \right|=\sqrt{\dfrac{16 \times 122}{{{\left( 61 \right)}^{2}}}} \\
 & \Rightarrow \left| z \right|=\dfrac{4\sqrt{122}}{61} \\
\end{align}$
Therefore \[\Rightarrow r=\dfrac{4\sqrt{122}}{61}\]
Now, we know that $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
Substituting the values we will get
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{-4}{61}}{\dfrac{44}{61}} \right)$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{-4}{44} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{-1}{11} \right) \\
 & \Rightarrow \theta =0.09 \\
\end{align}$
Now, the trigonometric representation of given complex number will be
$\Rightarrow z=r\left( \cos \theta +i\sin \theta \right)$
Now, substituting the values we will get
$\Rightarrow z=\dfrac{4\sqrt{122}}{61}\left( \cos 0.09-i\sin 0.09 \right)$
Hence above is the required trigonometric form.

Note: As the solution is lengthy please avoid calculation mistakes. To solve such type of questions students must know the trigonometric form or polar representation of complex numbers. The point to be noted is that the value of modulus is always positive.