Usando as equações de Cauchy-Riemann, prove:
\frac{d e^{a z}}{d z}=a e^{a z}

Passo 1

O enunciado pede para mostrar que a função
f(z)=e^{a z},\quad a\in\mathbb{C},
é holomorfa (satisfaça as equações de Cauchy–Riemann) e que sua derivada complexa é
f'(z)=a\,e^{a z}.

Passo 2

Escrevemos \displaystyle z=x+i y e \displaystyle a=\alpha+i\beta. Então
a\,z=(\alpha+i\beta)(x+i y)=(\alpha x-\beta y)+i(\beta x+\alpha y)
e
f(z)=e^{a z}=e^{\alpha x-\beta y}\bigl[\cos(\beta x+\alpha y)+i\sin(\beta x+\alpha y)\bigr].
Definimos
u(x,y)=e^{\alpha x-\beta y}\cos(\beta x+\alpha y),
v(x,y)=e^{\alpha x-\beta y}\sin(\beta x+\alpha y).

Passo 3

Calculamos as derivadas parciais de u e v:
u_x=\alpha e^{\alpha x-\beta y}\cos(\beta x+\alpha y)\;-\;\beta e^{\alpha x-\beta y}\sin(\beta x+\alpha y),
u_y=-\beta e^{\alpha x-\beta y}\cos(\beta x+\alpha y)\;-\;\alpha e^{\alpha x-\beta y}\sin(\beta x+\alpha y),
v_x=\beta e^{\alpha x-\beta y}\cos(\beta x+\alpha y)\;+\;\alpha e^{\alpha x-\beta y}\sin(\beta x+\alpha y),
v_y=-\alpha e^{\alpha x-\beta y}\cos(\beta x+\alpha y)\;+\;\beta e^{\alpha x-\beta y}\sin(\beta x+\alpha y).

Passo 4

Verificamos as equações de Cauchy–Riemann:
u_x = \alpha e^{\alpha x-\beta y}\cos(\cdots)-\beta e^{\alpha x-\beta y}\sin(\cdots) = v_y,
u_y = -\beta e^{\alpha x-\beta y}\cos(\cdots)-\alpha e^{\alpha x-\beta y}\sin(\cdots) = -\,v_x.
Como u_x=v_y e u_y=-v_x, a função é holomorfa.

Passo 5

Pela fórmula da derivada em termos de u_x e v_x, temos
f'(z)=u_x + i\,v_x =\bigl(\alpha+i\beta\bigr)e^{\alpha x-\beta y}\bigl[\cos(\beta x+\alpha y)+i\sin(\beta x+\alpha y)\bigr] = a\,e^{a z}.

Resposta

\frac{d}{dz}\,e^{a z}=a\,e^{a z}.

Exercícios relacionados