Question

How do you use DeMoivre's Theorem to simplify ${{\left( 3-2i \right)}^{8}}$?

Answer 1

This type of question is based on the concept of complex numbers and DeMoivre’s Theorem. We have to first consider the given function and compare with DeMoivre’s Theorem, that is “if$z=r\left( \cos \theta +i\sin \theta \right)$, then ${{z}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$”. By comparing DeMoivre's Theorem, we get $z=\left( 3-2i \right)$,$r\cos \theta =3$, $r\sin \theta =-2$ and n=8. We then have to find the value of r by squaring and adding $r\cos \theta =3$ and $r\sin \theta =-2$. And use the trigonometric identity, that is, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of r. then, substitute all the obtained values in ${{z}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$ to get the simplified form of ${{\left( 3-2i \right)}^{8}}$. We should also find the value of $\theta$.

Complete step-by-step solution:
According to the question, we are asked to find the simplified form of ${{\left( 3-2i \right)}^{8}}$ using DeMoivre’s Theorem.
We have been given the complex function ${{\left( 3-2i \right)}^{8}}$. ----------(1)
First, we have to consider ${{\left( 3-2i \right)}^{8}}$.
We know that, the DeMoivre’s Theorem is “if$z=r\left( \cos \theta +i\sin \theta \right)$, then ${{z}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$”.
Let us now compare this theorem with the given function.
We find that $z=\left( 3-2i \right)$ and n=8.
Here, ${{z}^{n}}={{\left( 3-2i \right)}^{8}}$.
On converting 3 and -2 to polar form, we get
$r\cos \theta =3$ and $r\sin \theta =-2$.
We need to find the value of r.
We have to do square $r\cos \theta =3$ and $r\sin \theta =-2$ on both sides.
Therefore, ${{\left( r\cos \theta \right)}^{2}}={{3}^{2}}$ and ${{\left( r\sin \theta \right)}^{2}}={{\left( -2 \right)}^{2}}$.
First consider ${{\left( r\cos \theta \right)}^{2}}={{3}^{2}}$.
By using the property ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$, we get
${{r}^{2}}{{\cos }^{2}}\theta ={{3}^{2}}$
Since ${{3}^{2}}=9$, we get
${{r}^{2}}{{\cos }^{2}}\theta =9$ ------------(2)
Now consider ${{\left( r\sin \theta \right)}^{2}}={{\left( -2 \right)}^{2}}$.
By using the property ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$, we get
${{r}^{2}}{{\sin }^{2}}\theta ={{\left( -2 \right)}^{2}}$
Since ${{\left( -2 \right)}^{2}}=4$, we get
${{r}^{2}}{{\sin }^{2}}\theta =4$ -----------(3)
Let us now add equation (2) and (3).
${{r}^{2}}{{\sin }^{2}}\theta +{{r}^{2}}{{\cos }^{2}}\theta =9+4$
On further simplifications, we get
${{r}^{2}}{{\sin }^{2}}\theta +{{r}^{2}}{{\cos }^{2}}\theta =13$
Here, we find that ${{r}^{2}}$ is common in the left-hand side. On taking ${{r}^{2}}$ out of the bracket, we get
${{r}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=13$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Using this trigonometric identity here, we get
${{r}^{2}}\left( 1 \right)=13$
$\Rightarrow {{r}^{2}}=13$
Now, take square root on both the sides of the expression.
$\Rightarrow \sqrt{{{r}^{2}}}=\sqrt{13}$
We know that $\sqrt{{{x}^{2}}}=x$. On using this property in the above expression, we get
$r=\sqrt{13}$
Now, we have to find $\theta$.
To find $\theta$, we have to divide $r\sin \theta =-2$ by $r\cos \theta =3$.
Therefore, $\dfrac{r\sin \theta }{r\cos \theta }=\dfrac{-2}{3}$.
Since r is common in both the numerator and denominator, we can cancel r.
$\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{-2}{3}$
We know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$. We get
$\tan \theta =\dfrac{-2}{3}$
To find $\theta$, we have to take ${{\tan }^{-1}}$ on both the sides of the obtained expression.
$\Rightarrow {{\tan }^{-1}}\left( \tan \theta \right)={{\tan }^{-1}}\left( \dfrac{-2}{3} \right)$
Using the identity ${{\tan }^{-1}}\left( \tan \theta \right)=\theta$ in the above expression, we get
$\theta ={{\tan }^{-1}}\left( \dfrac{-2}{3} \right)$.
Now, we have to find the simplified form of ${{\left( 3-2i \right)}^{8}}$ using DeMoivre’s Theorem.
Using the theorem, we get ${{z}^{n}}={{r}^{n}}\left( \cos n\theta +in\sin \theta \right)$
That is, ${{\left( 3-2i \right)}^{8}}={{\left( \sqrt{13} \right)}^{8}}\left( \cos 8\theta +i \sin 8\theta \right)$
We can write $\sqrt{13}={{13}^{\dfrac{1}{2}}}$ in the above equation.
$\Rightarrow {{\left( 3-2i \right)}^{8}}={{\left( {{13}^{\dfrac{1}{2}}} \right)}^{8}}\left( \cos 8\theta +i\sin 8\theta \right)$
Using the property ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{nm}}$ in the right-hand side of the equation, we get
$\Rightarrow {{\left( 3-2i \right)}^{8}}={{13}^{\dfrac{1}{2}\times 8}}\left( \cos 8\theta +i\sin 8\theta \right)$
On further simplification, we get
$\Rightarrow {{\left( 3-2i \right)}^{8}}={{13}^{4}}\left( \cos 8\theta +i\sin 8\theta \right)$.
Therefore, the simplified form of ${{\left( 3-2i \right)}^{8}}$ is ${{13}^{4}}\left( \cos 8\theta +i\sin 8\theta \right)$ where $\theta ={{\tan }^{-1}}\left( \dfrac{-2}{3} \right)$.

Note: We should use only DeMoivre’s Theorem as mentioned in this question. Then we do not forget to find the value of $\theta$ without which the solution is incomplete. Always convert the real part and imaginary part of the complex number to polar form at the first and then solve the rest. We should not make calculation mistakes based on sign conventions.