Question

How do you use De Moivre’s theorem to find $ {{\left( 1+i \right)}^{20}} $ in slandered form?

Answer 1

Hint: We explain the derivation of De Moivre’s theorem from Euler’s theorem. We state the theorems related to the power of $ {{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos \left( n\theta \right)+i\sin \left( n\theta \right) $ . The given equation $ \left( 1+i \right) $ is not in unit circle region. We convert it by dividing with its modulus value. Then we apply the theorems to final answer.

Complete step by step answer:
De Moivre’s theorem is actually a derivation of the Euler’s theorem. The theorem tells us
 $ {{e}^{i\theta }}=\cos \theta +i\sin \theta $ . Here $ i $ represents the imaginary value where $ i=\sqrt{-1} $ .
The equation represents a circle of the unit radius in the complex plane. It is also called Gauss’s plane.
Here X-axis is a regular real axis but the Y-axis is regarded as the imaginary axis.
We also have the indices theorem for $ {{e}^{i\theta }}=\cos \theta +i\sin \theta $ where
 $ {{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos \left( n\theta \right)+i\sin \left( n\theta \right) $ .
The given imaginary number $ \left( 1+i \right) $ doesn’t lie in the unit circle as $ \left| 1+i \right|=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2} $ .
We need to convert it into a unit circle equation to apply De Moivre’s theorem directly.
We can divide the equation $ \left( 1+i \right) $ by its norm $ \left| 1+i \right|=\sqrt{2} $ .
We get that $ \left( 1+i \right)=\sqrt{2}\left( \dfrac{1+i}{\sqrt{2}} \right)=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right) $ .
Now if $ z=\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right) $ then \[\left| z \right|=\sqrt{{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}}=1\].
We can apply De Moivre’s theorem directly for z where $ \left( 1+i \right)=\sqrt{2}z $ .
Therefore, \[{{\left( 1+i \right)}^{20}}={{\left( \sqrt{2}z \right)}^{20}}={{\left( \sqrt{2} \right)}^{20}}{{\left( z \right)}^{20}}\].
Now we solve both the constant and the imaginary part.
From indices theorem we know \[{{\left( \sqrt{2} \right)}^{20}}={{\left( {{2}^{\dfrac{1}{2}}} \right)}^{20}}={{2}^{\dfrac{1}{2}\times 20}}={{2}^{10}}\].
We know $ z=\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right)=\left[ \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right] $
Now, \[{{\left( 1+i \right)}^{20}}={{\left( \sqrt{2}z \right)}^{20}}={{\left( \sqrt{2} \right)}^{20}}{{\left( z \right)}^{20}}\].
We solve for \[{{\left( z \right)}^{20}}\] where \[{{\left( z \right)}^{20}}={{\left[ \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right]}^{20}}\].
Applying $ {{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos \left( n\theta \right)+i\sin \left( n\theta \right) $ we get,
\[\begin{align}
  & {{\left( z \right)}^{20}}={{\left[ \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right]}^{20}} \\
 & =\cos \left( 20\times \dfrac{\pi }{4} \right)+i\sin \left( 20\times \dfrac{\pi }{4} \right) \\
 & =\cos \left( 5\pi \right)+i\sin \left( 5\pi \right) \\
\end{align}\]
Now we solve for the angle $ 5\pi $ . We get \[\cos \left( 5\pi \right)+i\sin \left( 5\pi \right)=-1\].
Therefore, \[{{\left( z \right)}^{20}}=-1\].
We get \[{{\left( 1+i \right)}^{20}}=\left( -1 \right)\times {{2}^{10}}=-{{2}^{10}}\].

Note:
We need to remember that in the complex plane these coordinates are related in the same form as the real plane. For coordinates $ \left( x,y \right) $ , we have $ x=r\cos \theta $ and $ y=r\sin \theta $ . Here $ \theta $ is the angle of the joining line of the point in the unit circle with the origin.