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Complex Exponentials and Trigonometric functions
August 26th 2025
The value of the exponential function with an imaginary input is of interest.
e i x = ?
e^{ix} = ?
e i x = ? Using the well known series expansion of the exponential function:
e x = ∑ n = 0 ∞ x n n !
e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}
e x = n = 0 ∑ ∞ n ! x n Substitute in imaginary constant
e i x = ∑ n = 0 ∞ ( i ) n x n n !
e^{ix} = \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}
e i x = n = 0 ∑ ∞ n ! ( i ) n x n There is no direct way to simplify this expression as
( i ) n (i)^n ( i ) n
e i x = ∑ n = 0 ∞ ( i ) n x n n ! = ∑ n = 0 ∞ ( i ) 2 n x 2 n ( 2 n ) ! + ( i ) 2 n + 1 x 2 n + 1 ( 2 n + 1 ) ! = ∑ n = 0 ∞ ( i ) 4 n x 4 n ( 4 n ) ! + ( i ) 4 n + 1 x 4 n + 1 ( 4 n + 1 ) ! + ( i ) 4 n + 2 x 4 n + 2 ( 4 n + 2 ) ! + ( i ) 4 n + 3 x 4 n + 3 ( 4 n + 3 ) !
\begin{aligned}
e^{ix}
&= \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}\\
&= \sum_{n=0}^{\infty} \frac{(i)^{2n}x^{2n}}{(2n)!} + \frac{(i)^{2n+1}x^{2n+1}}{(2n+1)!}\\
&= \sum_{n=0}^{\infty} \frac{(i)^{4n}x^{4n}}{(4n)!} + \frac{(i)^{4n+1}x^{4n+1}}{(4n+1)!} + \frac{(i)^{4n+2}x^{4n+2}}{(4n+2)!} + \frac{(i)^{4n+3}x^{4n+3}}{(4n+3)!}
\end{aligned}
e i x = n = 0 ∑ ∞ n ! ( i ) n x n = n = 0 ∑ ∞ ( 2 n )! ( i ) 2 n x 2 n + ( 2 n + 1 )! ( i ) 2 n + 1 x 2 n + 1 = n = 0 ∑ ∞ ( 4 n )! ( i ) 4 n x 4 n + ( 4 n + 1 )! ( i ) 4 n + 1 x 4 n + 1 + ( 4 n + 2 )! ( i ) 4 n + 2 x 4 n + 2 + ( 4 n + 3 )! ( i ) 4 n + 3 x 4 n + 3 Now the powers of
i i i
i 4 n = i 4 n = 1 i 4 n + 1 = i 4 n i = i i 4 n + 2 = i 4 n i 2 = − 1 i 4 n + 3 = i 4 n i 3 = − i
\begin{aligned}
i^{4n} &= i^{4^n} = 1\\
i^{4n+1} &= i^{4n}i = i\\
i^{4n+2} &= i^{4n}i^2 = -1\\
i^{4n+3} &= i^{4n}i^3 = -i
\end{aligned}
i 4 n i 4 n + 1 i 4 n + 2 i 4 n + 3 = i 4 n = 1 = i 4 n i = i = i 4 n i 2 = − 1 = i 4 n i 3 = − i Back-substituting these into the above sum
e i x = ∑ n = 0 ∞ x 4 n ( 4 n ) ! + i x 4 n + 1 ( 4 n + 1 ) ! − x 4 n + 2 ( 4 n + 2 ) ! − i x 4 n + 3 ( 4 n + 3 ) !
e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!}+i\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+2}}{(4n+2)!}-i\frac{x^{4n+3}}{(4n+3)!}
e i x = n = 0 ∑ ∞ ( 4 n )! x 4 n + i ( 4 n + 1 )! x 4 n + 1 − ( 4 n + 2 )! x 4 n + 2 − i ( 4 n + 3 )! x 4 n + 3 Remember the power series for sine and cosine
sin ( x ) = ∑ n = 0 ∞ x 2 n + 1 ( − 1 ) n ( 2 n + 1 ) ! cos ( x ) = ∑ n = 0 ∞ x 2 n ( − 1 ) n ( 2 n ) !
\begin{aligned}
\sin(x) &=\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!}\\
\cos(x) &=\sum_{n=0}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!}
\end{aligned}
sin ( x ) cos ( x ) = n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 ( − 1 ) n = n = 0 ∑ ∞ ( 2 n )! x 2 n ( − 1 ) n Reorder series
e i x = ∑ n = 0 ∞ x 4 n ( 4 n ) ! − x 4 n + 2 ( 4 n + 2 ) ! + i x 4 n + 1 ( 4 n + 1 ) ! − i x 4 n + 3 ( 4 n + 3 ) !
e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!} + i \frac{x^{4n+1}}{(4n+1)!} - i\frac{x^{4n+3}}{(4n+3)!}
e i x = n = 0 ∑ ∞ ( 4 n )! x 4 n − ( 4 n + 2 )! x 4 n + 2 + i ( 4 n + 1 )! x 4 n + 1 − i ( 4 n + 3 )! x 4 n + 3 Split sums based on evens, and odds
e i x = [ ∑ n = 0 ∞ x 4 n ( 4 n ) ! − x 4 n + 2 ( 4 n + 2 ) ! ] + [ i ∑ n = 0 ∞ x 4 n + 1 ( 4 n + 1 ) ! − x 4 n + 3 ( 4 n + 3 ) ! ]
\begin{aligned}
e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!}}\right]\\
&+ \left[i\sum_{n=0}^{\infty} \frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right]
\end{aligned}
e i x = [ n = 0 ∑ ∞ ( 4 n )! x 4 n − ( 4 n + 2 )! x 4 n + 2 ] + [ i n = 0 ∑ ∞ ( 4 n + 1 )! x 4 n + 1 − ( 4 n + 3 )! x 4 n + 3 ] Notice the first sum is simply iterating over the even numbers, and the second is iterating over the odds. Simplify
e i x = [ ∑ n = 0 ∞ x 2 n ( − 1 ) n ( 2 n ) ! ] + [ i ∑ n = 0 ∞ x 2 n + 1 ( − 1 ) n ( 2 n + 1 ) ! ]
\begin{aligned}
e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{2n}(-1)^n}{(2n)!}}\right]\\
&+ \left[i\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!} \right]
\end{aligned}
e i x = [ n = 0 ∑ ∞ ( 2 n )! x 2 n ( − 1 ) n ] + [ i n = 0 ∑ ∞ ( 2 n + 1 )! x 2 n + 1 ( − 1 ) n ] Notice
e i x = cos x + i sin x
e^{ix} = \cos{x} + i\sin{x}
e i x = cos x + i sin x With Negatives
Recall these two properties of sine and cosine
sin ( − x ) = − sin ( x ) cos ( − x ) = cos ( x )
\begin{aligned}
\sin(-x) &= -\sin(x)\\
\cos(-x) &= \cos(x)\\
\end{aligned}
sin ( − x ) cos ( − x ) = − sin ( x ) = cos ( x ) With that, use the exponential function with a negative imaginary number
e − i x = cos x − i sin x
e^{-ix} = \cos{x} - i\sin{x}
e − i x = cos x − i sin x We can now solve for cosine and sine using the exponential function.
e − i x + e i x = 2 cos x
e^{-ix} + e^{ix} = 2\cos{x}
e − i x + e i x = 2 cos x
cos x = e i x + e − i x 2
\cos{x} = \frac{e^{ix} + e^{-ix}}{2}
cos x = 2 e i x + e − i x e i x − e − i x = 2 sin x
e^{ix} - e^{-ix} = 2\sin{x}
e i x − e − i x = 2 sin x
sin x = e i x − e − i x 2
\sin{x} = \frac{e^{ix} - e^{-ix}}{2}
sin x = 2 e i x − e − i x
This note is one of many taken during 2025