Answer.
\(e^{i\pi} = -1\text{,}\) \(e^{i\pi/2} = i\text{,}\) and \(e^{i\pi/4}=\sqrt{2}/2+i\sqrt{2}/2\text{.}\)
Skip to main content
Compute \(e^{i\pi}\text{,}\) \(e^{i\pi/2}\text{,}\) and \(e^{i\pi/4}\text{.}\)
Section A.3 Exponentiation of Complex Numbers
Subsection A.3.1 Definition Using Taylor Series
Complex numbers can be exponentiated. Let us begin by defining \(e^{bi}\) where \(b\) is a real number. Recall from basic calculus that if \(y\) is any real number then \(e^y\) has a Taylor series based at \(y=0\text{,}\)
\begin{equation}
1+y+y^2/2!+y^3/3!+ \cdots + y^k/k! + \cdots = \sum_{k=0}^{\infty} \frac{y^k}{k!}\tag{A.3.1}
\end{equation}
where \(k!=k(k-1)(k-2)\cdots 3\cdot 2\cdot 1\) and \(0!=1\text{.}\) The series in (A.3.1) converges to \(e^y\) for any real number \(y\text{.}\)
This can be used to define \(e^{bi}\text{,}\) by inserting \(y=bi\) into (A.3.1). It can be shown that the resulting series with complex terms
\begin{equation*}
1 + bi + (bi)^2/2! + (bi)^3/3! + \cdots + (bi)^k/k! + \cdots
\end{equation*}
converges to some complex number, for any real \(b\text{.}\) This complex number is what we will call \(e^{bi}\text{,}\) so we define
\begin{equation}
e^{bi} = 1 + bi + (bi)^2/2! + (bi)^3/3! + \cdots + (bi)^k/k! + \cdots\tag{A.3.2}
\end{equation}
although the series on the right hand side isn’t very explicit. In the next section we make it so.
Subsection A.3.2 Euler’s Formula
The series on the right in (A.3.2) can be made more explicit by noting that \(i^2=-1\text{,}\) \(i^3=-i\text{,}\) \(i^4=1\text{,}\) \(i^5=i\text{,}\) and so on. In short, the powers of \(i\) run through the sequence \(1\text{,}\) \(i\text{,}\) \(-1\text{,}\) \(-i\text{,}\) and then repeat with a cycle of length four. Based on this the series in (A.3.2) can be written as
\begin{equation*}
e^{bi} = 1 + bi - b^2/2! - ib^3/3! + b^4/4! + ib^5/5! + \cdots
\end{equation*}
Regroup terms (which is permitted for this infinite sum) to obtain
\begin{equation}
\begin{split}
e^{bi} = \amp \; \underbracket[1pt]{\left ( 1 - b^2/2! + b^4/4! - b^6/6! + \cdots \right )}_{\text{Re}(e^{bi})}\\
\amp + i\underbracket[1pt]{\left (b - b^3/3! + b^5/5! + \cdots \right )}_{\text{Im}(e^{bi})}.
\end{split}\tag{A.3.3}
\end{equation}
The series that defines the real part of \(e^{bi}\) on the right in (A.3.3) is just the Taylor series for \(\cos(b)\) from elementary calculus, and the series that defines the imaginary part is \(\sin(b)\text{.}\) We then have
\begin{equation}
e^{bi} = \cos(b) + i\sin(b)\tag{A.3.4}
\end{equation}
for any real number. This is known as Euler’s formula.
More generally we define \(e^z\) for any complex number \(z=a+bi\) by imposing the property \(e^{a+bi}=e^ae^{bi}\) (analogous to \(e^{x+y}=e^xe^y\) for real numbers \(x\) and \(y\)) to obtain
\begin{equation}
e^{a+bi} = e^ae^{bi} = e^a (\cos(b) + i\sin(b)) = e^a\cos(b) + ie^a\sin(b)\text{.}\tag{A.3.5}
\end{equation}
That is, the real part of \(e^{a+bi}\) is \(e^a\cos(b)\) and the imaginary part is \(e^a\sin(b)\text{.}\) With this definition complex exponentiation has many of the same algebraic properties as exponentiation for real numbers. In particular \(e^0=1\text{,}\) \(e^{w+z}=e^w e^z\text{,}\) and \(e^{-z}=1/e^z\text{.}\)
Reading Exercise A.3.1.
Subsection A.3.3 Sine and Cosine
\begin{equation*}
e^{-bi}=\cos(b)-i\sin(b)
\end{equation*}
since \(\sin(-b)=-\sin(b)\text{.}\) Adding this equation to (A.3.4) (left sides and right sides) and dividing by \(2\) shows that
\begin{equation*}
\cos(b) = \frac{e^{bi}+e^{-bi}}{2}\text{.}
\end{equation*}
Subtracting the equations and dividing by \(2i\) yields
\begin{equation*}
\sin(b) = \frac{e^{bi}-e^{-bi}}{2i}\text{.}
\end{equation*}
So the familiar trigonometric functions can be expressed using complex exponential functions.
