Question

How do you use De Moivre’s theorem to simplify ${(1 + i)^4}$

Answer 1

Hint : Here we have to find the value of ${(1 + i)^4}$ by using the theorem De Moivre’s theorem. Since the number is of the complex number where it contains both real part and imaginary part. We will use the concept of trigonometry also to find the values.

Complete step-by-step answer :
The given number is a complex number and it contains both the real part and imaginary part. The “ $i$ ” in the number represents the imaginary.
Now we will consider the trigonometry ratios and we will determine at what value of x sine and cosine has 1.
By the table of angles for the trigonometry ratios we have
$\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
The given question can be rewritten as
${(1 + i)^4} = {\left( {\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right)} \right)^4}$
We will write in the form of trigonometric ratios so we have
$\Rightarrow {\left( {\sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right)} \right)^4}$
To solve this, we use De moivre’s theorem, it states that
For any complex number x we have
${\left( {\cos x + i\sin x} \right)^n} = \cos (nx) + i\sin (nx)$
Where n is a positive integer and “ $i$ ” is the imaginary part $i = \sqrt { - 1}$ and also ${i^2} = - 1$
By using the De Moivre’s theorem
$\Rightarrow {\left( {\sqrt 2 } \right)^4}{\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)^4} \\\ \Rightarrow 4.\left( {\cos \left( {4\left( {\dfrac{\pi }{4}} \right)} \right) + i\sin \left( {4\left( {\dfrac{\pi }{4}} \right)} \right)} \right) \;$
On cancelling the 4
$\Rightarrow 4.\left( {\cos \pi + i\sin \pi } \right)$
The value of $\cos \pi$ is -1 and the value of $\sin \pi$ is 0. By substituting these values
$\Rightarrow 4.\left( { - 1 + i(0)} \right) \\\ \Rightarrow 4( - 1 + 0) \\\ \Rightarrow 4( - 1) \\\ \Rightarrow - 4 \;$
So, the correct answer is “-4”.

Note : A complex number is a combination of real number and the imaginary number where $i$ represents the imaginary. To solve or simplify the complex number De Moivre’s theorem. It is stated as For any complex number x we have ${\left( {\cos x + i\sin x} \right)^n} = \cos (nx) + i\sin (nx)$ where n is a positive integer and “ $i$ ” is the imaginary part $i = \sqrt { - 1}$ and also ${i^2} = - 1$ . The trigonometry ratios values for the standard angles are used to simplify further.