Question

How do you use De Moivre’s theorem to simplify $ {{\left( 2+2i \right)}^{6}} $ ?

Answer 1

Hint:
We start solving the problem by making use of the result $ {{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}} $ for the given complex number. We then make the necessary calculations to the complex number present inside the exponent and then make use of the fact that \[\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\] and \[\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\] to proceed through the problem. We then make use of the De Moivre’s theorem that if a complex number is in the form of $ {{\left( \cos x+i\sin x \right)}^{n}} $ , then it is equal to $ \left( \cos \left( nx \right)+i\sin \left( nx \right) \right) $ to proceed further through the problem. We then make use of the results \[\sin \left( \dfrac{3\pi }{2} \right)=-1\], \[\cos \left( \dfrac{3\pi }{2} \right)=0\] and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to apply the De Moivre’s theorem to simplify $ {{\left( 2+2i \right)}^{6}} $ .
Let us assume $ Z={{\left( 2+2i \right)}^{6}} $ .
 $ \Rightarrow Z={{\left( 2\left( 1+i \right) \right)}^{6}} $ ---(1).
We know that $ {{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}} $ . Let us use this result in equation (1).
 $ \Rightarrow Z={{2}^{6}}{{\left( 1+i \right)}^{6}} $ .
 $ \Rightarrow Z=32{{\left( \left( \sqrt{2} \right)\left( \dfrac{1+i}{\sqrt{2}} \right) \right)}^{6}} $ ---(2).
We know that $ {{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}} $ . Let us use this result in equation (2).
 $ \Rightarrow Z=32{{\left( \sqrt{2} \right)}^{6}}{{\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right)}^{6}} $ .
 $ \Rightarrow Z=32\left( 8 \right){{\left( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right)}^{6}} $ ---(3).
We know that \[\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\] and \[\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\]. Let us use these results in equation (3).
 $ \Rightarrow Z=256{{\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right)}^{6}} $ ---(4).
Now, let us recall the definition of De Moivre’s theorem. We know that De Moivre’s theorem states that if a complex number is in the form of $ {{\left( \cos x+i\sin x \right)}^{n}} $ , then it is equal to $ \left( \cos \left( nx \right)+i\sin \left( nx \right) \right) $ . Let us use this result in equation (4).
 $ \Rightarrow Z=256\left( \cos \left( 6\times \dfrac{\pi }{4} \right)+i\sin \left( 6\times \dfrac{\pi }{4} \right) \right) $ .
 $ \Rightarrow Z=256\left( \cos \left( \dfrac{3\pi }{2} \right)+i\sin \left( \dfrac{3\pi }{2} \right) \right) $ ---(5).
We know that \[\sin \left( \dfrac{3\pi }{2} \right)=-1\] and \[\cos \left( \dfrac{3\pi }{2} \right)=0\]. Let us use these results in equation (3).
 $ \Rightarrow Z=256\left( 0+i\left( -1 \right) \right) $ .
 $ \Rightarrow Z=-256i $ .
So, we have found the value of $ {{\left( 2+2i \right)}^{6}} $ as $ -256i $ .

Note:
 We should perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problems, we first try to convert the given complex number into the form $ {{r}^{n}}{{\left( \cos \theta +i\sin \theta \right)}^{n}} $ to make use of the De Moivre’s theorem. We can also solve this problem by converting the given complex number into the Euler’s form and then making the necessary calculations to get the required answer. Similarly, we can expect the problems to find the Euler’s form of the obtained result.