Question

How do you use De Moivre’s theorem to express ${(1 + i)^8}$ ?

Answer 1

Hint: We will first start by mentioning De Moivre’s Theorem. Then apply the theorem, and note all the values of $n$ for which we will solve. Then evaluate all the values for different values of $n$ and hence, evaluate the fourth roots of $i$. Here we will be using $r = \sqrt {{a^2} + {b^2}} $ and $\theta = \arctan \left( {\dfrac{b}{a}} \right)$.

Complete step-by-step answer:
Here we will start by using the De Moivre’s Theorem.
According to the theorem, $z = r{e^{i\theta }} = r(\cos \theta + i\sin \theta )$.
We will now convert the given term from complex form to trig form.
$a + bi \to r(\cos (\theta ) + i\sin (\theta ))$
With the help of $r = \sqrt {{a^2} + {b^2}} $ and $\theta = \arctan \left( {\dfrac{b}{a}} \right)$.
Here, we have the number, ${(1 + i)^8}$
So, now we compare and evaluate the values of the term.
 \[
  r = \sqrt {{1^2} + {1^2}} = \sqrt 2 \\
  \theta = \arctan \left( {\dfrac{1}{1}} \right) = \dfrac{\pi }{4} \\
  z = \sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right) \;
 \]
So , now here \[z = \sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right)\] is the trigonometric form.
Now, we will apply the De Moivre’s Theorem.
 \[
  z = \sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right) \\
  {z^8} = {\left( {\sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right)} \right)^8} \\
  {z^8} = {\sqrt 2 ^8}\left( {\cos \left( {8\dfrac{\pi }{4}} \right) + i\sin \left( {8\dfrac{\pi }{4}} \right)} \right) \\
  {z^8} = 16\left( {\cos \left( {2\pi } \right) + i\sin \left( {2\pi } \right)} \right) \\
  {z^8} = 16 \;
 \]
So, the correct answer is “$ {z^8} = 16 $”.

Note: Complex numbers is a number that can be expressed in the form of $a + ib$, where $a$ and $b$ are real numbers, and $i$ represents the imaginary unit, satisfying the equation ${i^2} = - 1$. Because no real number satisfies this equation, $i$ is called an imaginary number. Complex numbers allow solutions to certain equations that have no solutions in real numbers. The idea is to extend the real numbers with an intermediate $i$ which is also called an imaginary unit taken to satisfy the relation ${i^2} = - 1$, so that solutions to equations like the preceding one can be found.