4.9 Elementary Functions
The Exponential Function
The function \(\boxed {e^z = e^x\cdot e^{iy}}\) or \(\boxed {e^z = e^x\cos y + ie^x\sin y}\) is called the complex exponential function.
Theorem 4.57. If \(z_1\) and \(z_2\) are complex numbers, then
- 1.
- \(e^0 \neq 0, \quad e^0 = 1\)
- 2.
- \(e^{z_1 + z_2} = e^{x_1+x_2}\big (\cos (y_1 + y_2) + i\sin (y_1+ y_2)\big )\)
\(e^{z_1}\cdot e^{z_2} = e^{x_1 + x_2} \big [\cos (y_1 + y_2) + i\sin (y_1 + y_2)\big ]\)
- 3.
- If \(x\) is a real number then \(\quad e^{x + i0} = e^x (\cos (0) + i\sin (0)) = e^x\)
- 4.
- \(\left |e^z\right | = e^x\)
- 5.
- \(e^{iy} = \cos y + i\sin y\)
- 6.
- \(\frac {e^{z_1}}{e^{z_2}} = e^{z_1 -z_2}\)
- 7.
- \(\Big (e^z\Big )^n = e^{nz}\quad , \quad n \in \mathbb {Z}\)
- 8.
- \(\overline {e^z} = e^{\overline {z}}\)
- 9.
- \(e^z = \alpha :\quad \) This equation has infinitely many solutions \(z\),\(\quad \alpha \neq 0,\,\alpha \in \mathbb {C}\)
Example 4.58. \(e^z= 2 + 2i\) \begin {align*} 2 + 2i & = 2(1 + i) = 2\sqrt {2}\Big (\frac {1}{\sqrt {2}} + i \frac {1}{\sqrt {2}}\Big )\\\\ & = 2\sqrt {2}\Big (\cos \Big (\frac {\pi }{4}\Big ) + i \sin \Big (\frac {\pi }{4}\Big )\Big )\\\\ & = 2\sqrt {2}\Big (\cos \Big (\frac {\pi }{4} + 2n\pi \Big ) + i \sin \Big (\frac {\pi }{4} + 2n\pi \Big )\Big )\qquad n\in \mathbb {Z} \end {align*}
In the form \(e^x(\cos y + i\sin y),\quad \) take \(e^x = 2\sqrt {2}\) i.e take \(x = \ln (2\sqrt {2})\)
Take \(y = 2n\pi + \frac {\pi }{4}\,\) then \[e^z = \exp (\ln (2\sqrt {2}) + i (2n + 1/4)\pi = 2 + 2i \,, \qquad n\in \mathbb {Z}\] - 10.
- \(e^z\) is an entire function
\(f'(z) = (e^z)' = e^z = f(z)\)
Recall \((f = u + iv \) then \(f'(z) = u_x + iv_x)\) \begin {align*} e^z & = e^x \cos y + i e^x\sin y\\\ (e^z)' & = e^x \cos y + ie^x \sin y = e^z \end {align*}
- 11.
- \(e^{z + (2\pi i n)} = e^x\big [\cos (2\pi n + y ) + i \sin (2\pi n + y) \big ] = e^z\quad \) So \(e^z\) has a period \(2\pi i\)
Properties of Exponential Functions
Caution: [These do not hold for real exponential functions]
- 1.
- \(e^z\) may be a negative real number.
- 2.
- \(e^z\) is periodic with pure imaginary period \(2\pi i\).
- 3.
- For a given \(w_0\neq 0\) there may be many values of \(z\) with \(e^z = w_0\); that is, the inverse of the exponential function is many-valued.
Proof.
(1)
When \(y = n\pi \) with \(n\in \mathbb {Z}\), \begin {align*} e^z &= e^x \cos y + ie^x\sin y\\ &= e^x(-1)^n + i0\\ &= (-1)^ne^x = \begin {cases} e^x & \text {for}\ n = 2m\\ -e^x & \text {for}\ n = 2m + 1 \end {cases} \end {align*}
that is, \(e^{(2m + 1)\pi i} = -1\) and \(e^{2m\pi i} = 1\) for \(m\in \mathbb {Z}\). So \(e^z\) does take negative values, which \(e^x\) never does for real \(x\).
(2)
From \(e^{2m\pi i} = 1\) we get \[e^{z + 2\pi i} = e^z\cdot e^{2\pi i} = e^z\cdot 1 = e^z ,\] so \(e^z\) is periodic with period \(2\pi i\).
(3)
For \(w_0 = \left |w_0\right |e^{i \operatorname {Arg}(w_0)} \neq 0\) the equation \[e^x \cdot e^{iy} = w_0 = \left |w_0\right |e^{i\operatorname {Arg}(w_0)}\] gives \(e^x = \left |w_0\right |\) and \(y = \operatorname {Arg}(w_0) + 2n\pi \), so \(z = x+iy\) with \(x = \ln \left |w_0\right |\) and \(y = \operatorname {Arg}(w_0) + 2n\pi \) solves \(e^z = w_0\). Since \(n\) ranges over \(\mathbb {Z}\) there are infinitely many solutions. □
Solution. Since \(1+i\) has modulus \(\left |1+i\right | = \sqrt {2}\) and principal argument \(\operatorname {Arg}(1+i) = \frac {\pi }{4}\), \[e^x \cdot e^{iy} = e^z = 1 + i = \sqrt {2}\, e^{i\pi /4},\] so that \[e^x = \sqrt {2},\qquad y = \frac {\pi }{4} + 2n\pi ,\] that is \[x = \ln \sqrt {2} = \frac {\ln 2}{2},\qquad y = \Big (\frac {1}{4} + 2n\Big )\pi ,\qquad n\in \mathbb {Z}.\] Hence \[z = \frac {\ln 2}{2} + i\Big (\frac {1}{4}+ 2n\Big )\pi ,\] which is not unique, precisely because of the periodicity in property (2).
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