Chapter I, Problem 49E

If u(x) = f(x) + ig(x) is a complex-valued function of a real variable x and the real and imaginary parts f(x) and g(x) are differentiable functions of x, then the derivative of u is defined to be u′(x) = f′(x) + ig′(x). Use this together with Equation 7 to prove that if F(x) = e^rx, then F′(x) = re^rx when r = a + bi is a complex number.

To determine

To prove: If $F\left(x\right)=e^{rx}$ then, $F^{\prime }\left(x\right)=r{e}^{rx}$ where $r=a+bi$.

Explanation of Solution

Formula used:

Euler’s formula:

$e^{iy}=\cos y+i\sin y$

Calculation:

It is given that, if $u\left(x\right)=f\left(x\right)+ig\left(x\right)$ is a complex valued function then, the derivative of the function $u\left(x\right)$ is $u^{\prime }\left(x\right)=f^{\prime }\left(x\right)+i{g}^{\prime }\left(x\right)$.

Consider the function $F\left(x\right)=e^{rx}$ and rewrite the function by the use of Euler’s formula.

$\begin{array}{c}F\left(x\right)=e^{rx}\\ =e^{\left(a+bi\right)x}\\ =e^{ax+bxi}\\ =e^{ax}\left(\cos bx+i\sin bx\right)\end{array}$

That is, the function is $F\left(x\right)=e^{ax}\cos bx+i\left(e^{ax}\sin bx\right)$.

Obtain the derivative of the function $F\left(x\right)=e^{ax}\cos bx+i\left(e^{ax}\sin bx\right)$.

$\begin{array}{c}{F}^{\prime }\left(x\right)={\left(e^{ax}\cos bx\right)}^{\prime }+i{\left(e^{ax}\sin bx\right)}^{\prime }\\ =\left(e^{ax}\frac{d}{dx}\left(\cos bx\right)+\cos bx\frac{d}{dx}\left(e^{ax}\right)\right)+i\left(e^{ax}\frac{d}{dx}\left(\sin bx\right)+\sin bx\frac{d}{dx}\left(e^{ax}\right)\right)\\ =\left(a{e}^{bx}\cos bx-b{e}^{ax}\sin bx\right)+i\left(a{e}^{ax}\sin bx+b{e}^{ax}\cos bx\right)\\ =a{e}^{ax}\left(\cos bx+i\sin bx\right)+b{e}^{ax}\left(-\sin bx+i\cos bx\right)\end{array}$

Further simplified as,

$\begin{array}{c}{F}^{\prime }\left(x\right)=a\left[e^{ax}\left(\cos bx+i\sin bx\right)\right]+b{e}^{ax}\left(\left(i^2\right)\sin bx+i\cos bx\right)\left[\because i^2=1\right]\\ =a{e}^{rx}+ib{e}^{ax}\left(\cos bx+i\sin bx\right)\\ =a{e}^{rx}+ib{e}^{rx}\left[\because e^{ax}\left(\cos bx+i\sin bx\right)=e^{rx}\right]\\ =e^{rx}\left(a+ib\right)\\ =r{e}^{rx}\end{array}$

That is, the derivative $F^{\prime }\left(x\right)=r{e}^{rx}$.

Hence the proof.