How do you divide $\dfrac{3i-7}{2i-1}$ in trigonometric form?
Answer
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Hint: To divide the given complex number in trigonometric form we are going to first convert each of the complex numbers written in the numerator and denominator in the trigonometric form. Let us assume a complex number say $a+ib$ then the trigonometric form of this complex number is equal to $\sqrt{{{a}^{2}}+{{b}^{2}}}\left( \cos \theta +i\sin \theta \right)$ and $\theta $ is calculated by the formula $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$. This is the way we are going to convert the given complex number into trigonometric form and then we simplify.
Complete step-by-step solution:
The complex number given in the above problem is as follows:
$\dfrac{3i-7}{2i-1}$
Writing numerator of the above complex number in the form of trigonometric form as follows:
$\Rightarrow 3i-7$
Now, let us assume the argument of the above complex number is ${{\theta }_{1}}$ and writing the above complex number in the trigonometric form using $\sqrt{{{a}^{2}}+{{b}^{2}}}\left( \cos \theta +i\sin \theta \right)$ we get,
$\begin{align}
& =\sqrt{9+49}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\
& =\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\
\end{align}$
Similarly, we are going to write the trigonometric form of the complex number written in the denominator.
$\begin{align}
& =2i-1 \\
& =\sqrt{4+1}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\
& =\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\
\end{align}$
Now, substituting the above trigonometric form in the given complex number we get,
$\Rightarrow \dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}$
Multiplying and dividing the above fraction by $\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)$ we get,
$\begin{align}
& \Rightarrow \dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}\times \dfrac{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)} \\
& =\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}\left( {{\cos }^{2}}{{\theta }_{2}}-{{i}^{2}}{{\sin }^{2}}{{\theta }_{2}} \right)} \\
\end{align}$
We know that ${{i}^{2}}=-1$ so using this formula in the above we get,
$=\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}\left( {{\cos }^{2}}{{\theta }_{2}}+{{\sin }^{2}}{{\theta }_{2}} \right)}$
We also know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so applying this trigonometric property in the above we get,
\[\begin{align}
& =\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}} \\
& =\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{1}}\cos {{\theta }_{2}}-\cos {{\theta }_{1}}\sin {{\theta }_{2}} \right) \right) \\
\end{align}\]
We know the trigonometric identity i.e. \[\sin {{\theta }_{1}}\cos {{\theta }_{2}}-\cos {{\theta }_{1}}\sin {{\theta }_{2}}=\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\] so using this identity in the above we get,
\[=\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\]
Also, we know that $\cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}}=\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ so using this identity in the above we get,
\[=\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\]
Now, let us find the values of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ as follows:
Using the complex number written in the numerator of the above complex number:
$\Rightarrow 3i-7$
$\tan {{\theta }_{1}}=-\dfrac{3}{7}$
And we are going to write the tangent of the argument of the other complex number i.e. $2i-1$ we get,
$\begin{align}
& \Rightarrow \tan {{\theta }_{2}}=-\dfrac{2}{1} \\
& \Rightarrow \tan {{\theta }_{2}}=-2 \\
\end{align}$
Now, writing the tangent of subtraction of ${{\theta }_{1}}\And {{\theta }_{2}}$ we get,
\[\tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{\tan {{\theta }_{1}}-\tan {{\theta }_{2}}}{1+\tan {{\theta }_{1}}\tan {{\theta }_{2}}}\]
Substituting the values of \[\tan {{\theta }_{1}}\And \tan {{\theta }_{2}}\] from the above we get,
\[\begin{align}
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{-\dfrac{3}{7}-\left( -2 \right)}{1+\dfrac{3}{7}\left( 2 \right)} \\
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{\dfrac{-3+14}{7}}{\dfrac{7+6}{7}} \\
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{11}{13} \\
\end{align}\]
From the triangle, we know that $\tan \theta =\dfrac{P}{B}$ so comparing this form with the above tangent we get,
\[\Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{11}{13}=\dfrac{P}{B}\]
The hypotenuse (H) is calculated as follows:
$\begin{align}
& H=\sqrt{{{P}^{2}}+{{B}^{2}}} \\
& \Rightarrow H=\sqrt{{{11}^{2}}+{{13}^{2}}} \\
& \Rightarrow H=\sqrt{121+169} \\
& \Rightarrow H=\sqrt{290} \\
\end{align}$
Now, we are going to find the value of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ as follows:
$\begin{align}
& \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{B}{H}=\dfrac{13}{\sqrt{290}}; \\
& \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{P}{H}=\dfrac{11}{\sqrt{290}} \\
\end{align}$
Substituting these values of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ in \[\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\] we get,
\[\Rightarrow \dfrac{\sqrt{58}}{\sqrt{5}}\left( \dfrac{13}{\sqrt{290}}+i\left( \dfrac{11}{\sqrt{290}} \right) \right)\]
Taking $\sqrt{290}$ as common from the denominator and we get,
$\begin{align}
& =\dfrac{\sqrt{58}}{\sqrt{5}\left( \sqrt{290} \right)}\left( 13+11i \right) \\
& =\dfrac{\sqrt{29\times 2}}{\left( \sqrt{29\times 5\times 2\times 5} \right)}\left( 13+11i \right) \\
& =\dfrac{1}{5}\left( 13+11i \right) \\
\end{align}$
Hence, we have divided the given complex number using trigonometric form and the answer is as follows: $\dfrac{1}{5}\left( 13+11i \right)$
Note: The mistake that could be possible in the above problem is that while writing the values of $\tan {{\theta }_{1}}\And \tan {{\theta }_{2}}$, make sure you have properly put the negative sign wherever it is required just like we did in the above problem.
$\tan {{\theta }_{1}}=-\dfrac{3}{7}$
$\Rightarrow \tan {{\theta }_{2}}=-2$
Generally, we forgot to put the signs according to the complex number so make sure you won’t make this mistake.
Complete step-by-step solution:
The complex number given in the above problem is as follows:
$\dfrac{3i-7}{2i-1}$
Writing numerator of the above complex number in the form of trigonometric form as follows:
$\Rightarrow 3i-7$
Now, let us assume the argument of the above complex number is ${{\theta }_{1}}$ and writing the above complex number in the trigonometric form using $\sqrt{{{a}^{2}}+{{b}^{2}}}\left( \cos \theta +i\sin \theta \right)$ we get,
$\begin{align}
& =\sqrt{9+49}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\
& =\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\
\end{align}$
Similarly, we are going to write the trigonometric form of the complex number written in the denominator.
$\begin{align}
& =2i-1 \\
& =\sqrt{4+1}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\
& =\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\
\end{align}$
Now, substituting the above trigonometric form in the given complex number we get,
$\Rightarrow \dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}$
Multiplying and dividing the above fraction by $\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)$ we get,
$\begin{align}
& \Rightarrow \dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)}{\sqrt{5}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right)}\times \dfrac{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)} \\
& =\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}\left( {{\cos }^{2}}{{\theta }_{2}}-{{i}^{2}}{{\sin }^{2}}{{\theta }_{2}} \right)} \\
\end{align}$
We know that ${{i}^{2}}=-1$ so using this formula in the above we get,
$=\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}\left( {{\cos }^{2}}{{\theta }_{2}}+{{\sin }^{2}}{{\theta }_{2}} \right)}$
We also know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so applying this trigonometric property in the above we get,
\[\begin{align}
& =\dfrac{\sqrt{58}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right)\left( \cos {{\theta }_{2}}-i\sin {{\theta }_{2}} \right)}{\sqrt{5}} \\
& =\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}} \right)-{{i}^{2}}\left( \sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)+i\left( \sin {{\theta }_{1}}\cos {{\theta }_{2}}-\cos {{\theta }_{1}}\sin {{\theta }_{2}} \right) \right) \\
\end{align}\]
We know the trigonometric identity i.e. \[\sin {{\theta }_{1}}\cos {{\theta }_{2}}-\cos {{\theta }_{1}}\sin {{\theta }_{2}}=\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\] so using this identity in the above we get,
\[=\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos {{\theta }_{1}}\left( \cos {{\theta }_{2}} \right)+\left( \sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\]
Also, we know that $\cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}}=\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ so using this identity in the above we get,
\[=\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\]
Now, let us find the values of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ as follows:
Using the complex number written in the numerator of the above complex number:
$\Rightarrow 3i-7$
$\tan {{\theta }_{1}}=-\dfrac{3}{7}$
And we are going to write the tangent of the argument of the other complex number i.e. $2i-1$ we get,
$\begin{align}
& \Rightarrow \tan {{\theta }_{2}}=-\dfrac{2}{1} \\
& \Rightarrow \tan {{\theta }_{2}}=-2 \\
\end{align}$
Now, writing the tangent of subtraction of ${{\theta }_{1}}\And {{\theta }_{2}}$ we get,
\[\tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{\tan {{\theta }_{1}}-\tan {{\theta }_{2}}}{1+\tan {{\theta }_{1}}\tan {{\theta }_{2}}}\]
Substituting the values of \[\tan {{\theta }_{1}}\And \tan {{\theta }_{2}}\] from the above we get,
\[\begin{align}
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{-\dfrac{3}{7}-\left( -2 \right)}{1+\dfrac{3}{7}\left( 2 \right)} \\
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{\dfrac{-3+14}{7}}{\dfrac{7+6}{7}} \\
& \Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{11}{13} \\
\end{align}\]
From the triangle, we know that $\tan \theta =\dfrac{P}{B}$ so comparing this form with the above tangent we get,
\[\Rightarrow \tan \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{11}{13}=\dfrac{P}{B}\]
The hypotenuse (H) is calculated as follows:
$\begin{align}
& H=\sqrt{{{P}^{2}}+{{B}^{2}}} \\
& \Rightarrow H=\sqrt{{{11}^{2}}+{{13}^{2}}} \\
& \Rightarrow H=\sqrt{121+169} \\
& \Rightarrow H=\sqrt{290} \\
\end{align}$
Now, we are going to find the value of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ as follows:
$\begin{align}
& \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{B}{H}=\dfrac{13}{\sqrt{290}}; \\
& \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)=\dfrac{P}{H}=\dfrac{11}{\sqrt{290}} \\
\end{align}$
Substituting these values of $\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)\And \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right)$ in \[\dfrac{\sqrt{58}}{\sqrt{5}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\left( \sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \right)\] we get,
\[\Rightarrow \dfrac{\sqrt{58}}{\sqrt{5}}\left( \dfrac{13}{\sqrt{290}}+i\left( \dfrac{11}{\sqrt{290}} \right) \right)\]
Taking $\sqrt{290}$ as common from the denominator and we get,
$\begin{align}
& =\dfrac{\sqrt{58}}{\sqrt{5}\left( \sqrt{290} \right)}\left( 13+11i \right) \\
& =\dfrac{\sqrt{29\times 2}}{\left( \sqrt{29\times 5\times 2\times 5} \right)}\left( 13+11i \right) \\
& =\dfrac{1}{5}\left( 13+11i \right) \\
\end{align}$
Hence, we have divided the given complex number using trigonometric form and the answer is as follows: $\dfrac{1}{5}\left( 13+11i \right)$
Note: The mistake that could be possible in the above problem is that while writing the values of $\tan {{\theta }_{1}}\And \tan {{\theta }_{2}}$, make sure you have properly put the negative sign wherever it is required just like we did in the above problem.
$\tan {{\theta }_{1}}=-\dfrac{3}{7}$
$\Rightarrow \tan {{\theta }_{2}}=-2$
Generally, we forgot to put the signs according to the complex number so make sure you won’t make this mistake.
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