De Moivre's Formula/Positive Integer Index
Theorem
Let $z \in \C$ be a complex number expressed in polar form:
- $z = r \paren {\cos x + i \sin x}$
Then:
- $\forall n \in \Z_{>0}: \paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \map \sin {n x} }$
Corollary
- $\forall n \in \Z_{>0}: \paren {\cos x + i \sin x}^n = \map \cos {n x} + i \map \sin {n x}$
Proof 1
Proof by induction:
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$
$\map P 1$ is the case:
- $\paren {r \paren {\cos x + i \sin x} }^1 = r^1 \paren {\map \cos {1 x} + i \, \map \sin {1 x} }$
which is trivially true.
Basis for the Induction
$\map P 2$ is the case:
- $\paren {r \paren {\cos x + i \sin x} }^2 = r^2 \paren {\map \cos {n x} + i \, \map \sin {2 x} }$
From Product of Complex Numbers in Polar Form, we have:
- $r_1 \paren {\cos x_1 + i \sin x_1 } r_2 \paren {\cos x_2 + i \sin x_2} = r_1 r_2 \paren {\map \cos {x_1 + x_2} + i \, \map \sin {x_1 + x_2} }$
Setting $r_1 = r_2 = r$ and $x_1 = x_2 = x$ gives the result.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\paren {r \paren {\cos x + i \sin x} }^k = r^k \paren {\map \cos {k x} + i \, \map \sin {k x} }$
Then we need to show:
- $\paren {r \paren {\cos x + i \sin x} }^{k + 1} = r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }$
Induction Step
This is our induction step:
\(\ds \paren {r \paren {\cos x + i \sin x} }^{k + 1}\) | \(=\) | \(\ds \paren {r \paren {\cos x + i \sin x} }^k \paren {r \paren {\cos x + i \sin x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds r^k \paren {\map \cos {k x} + i \, \map \sin {k x} } \paren {r \paren {\cos x + i \sin x} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }\) | Product of Complex Numbers in Polar Form |
Hence, by induction, for all $n \in \Z_{> 0}$:
- $\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$
$\blacksquare$
Proof 2
From Product of Complex Numbers in Polar Form: General Result:
- $z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$
Setting $z_1 = z_2 = \cdots = z_n = r \paren {\cos x + i \sin x}$ gives the result.
Also known as
De Moivre's Theorem.
Source of Name
This entry was named for Abraham de Moivre.
Sources
- 1957: E.G. Phillips: Functions of a Complex Variable (8th ed.) ... (previous) ... (next): Chapter $\text I$: Functions of a Complex Variable: $\S 3$. Geometric Representation of Complex Numbers
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): mathematical induction: $\text {(iii)}$