Prove that [latex](a) (d/dz)e^z = e^z , (b) (d/dz)e^{az} = ae^{az}[/latex] where a is any constant.

By definition, [latex] w=e^z=e^{x+i y}=e^x(\cos y+i \sin y)=u+i v[/latex] or [latex]u=e^x \cos y, v=e^x \sin y[/latex]. Since [latex]\partial u / \partial x=e^x \cos y=\partial v / \partial y[/latex] and [latex]\partial v / \partial x=e^x \sin y=-(\partial u / \partial y)[/latex], the Cauchy-Riemann equations are satisfied. Then, by Problem 3.5, the required derivative exists and is equal to [latex]\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=-i \frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}=e^x \cos y+i e^x \sin y=e^z[/latex] (b) Let [latex]w=e^\zeta[/latex] where [latex]\zeta=a z[/latex]. Then, by part (a) and Problem 3.39 , [latex]\frac{d}{d z} e^{a z}=\frac{d}{d z} e^\zeta=\frac{d}{d \zeta} e^\zeta \cdot \frac{d \zeta}{d z}=e^\zeta \cdot a=a e^{a z}[/latex] We can also proceed as in part (a).

Question 3.11: Prove that (a) (d/dz)e^z = e^z , (b) (d/dz)e^az = ae^az wher...

Schaum’s Outline of Complex Variables

Prove that $(a) (d/dz)e^z = e^z , (b) (d/dz)e^{az} = ae^{az}$ where a is any constant.

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By definition, $w=e^z=e^{x+i y}=e^x(\cos y+i \sin y)=u+i v$ or $u=e^x \cos y, v=e^x \sin y$ .
Since $\partial u / \partial x=e^x \cos y=\partial v / \partial y$ and $\partial v / \partial x=e^x \sin y=-(\partial u / \partial y)$ , the Cauchy-Riemann equations are satisfied. Then, by Problem 3.5, the required derivative exists and is equal to

$\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=-i \frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}=e^x \cos y+i e^x \sin y=e^z$

(b) Let $w=e^\zeta$ where $\zeta=a z$ . Then, by part (a) and Problem 3.39,