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hotvette

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## Summary:

- Clarify statement in Complex Variables (S. Fisher) about magnitude of a line integral exp(iz) over a semicircle

## Main Question or Discussion Point

Not homework, just trying to understand a statement in the book. On page 158 in Fisher, the following statement is made:

In these applications of the Residue Theorem, we often need to estimate the magnitude of the line integral of [itex] e^{iz} [/itex] over the semicircle [itex]= Re^{i\theta}, \; 0 \le \theta \le \pi[/itex]. We note that:

[tex]

\left\vert e^{iz} \right\vert = e^{\text{Re}(iz)} = e^{-R \sin(\theta)}

[/tex]

To understand the above, I did the following.

\begin{align}

&z = Re^{i\theta} = R \cos(\theta) + i R \sin(\theta)

\\

&iz = i R \cos(\theta) - R \sin(\theta)

\\

&e^{iz} = e^{i R \cos(\theta) - R \sin(\theta)} = e^{i R \cos(\theta)} e^{-R \sin(\theta)}

\\

&\left\vert e^{iz} \right\vert = \left\vert e^{i R \cos(\theta)} e^{-R \sin(\theta)} \right\vert

= \left\vert e^{i R \cos(\theta)} \right\vert \; \left\vert e^{-R \sin(\theta)} \right\vert

\\

&e^{i R \cos(\theta)} = e^{ix} && \text{where $x = R \cos(\theta)$, real}

\\

\therefore \, &\left\vert e^{i R \cos(\theta)} \right\vert = 1

&& \cos^2 x + \sin^2 x= 1

\\

&\left\vert e^{-R \sin(\theta)} \right\vert = e^{-R \sin(\theta)}

&& \text{since } e^{-R \sin(\theta)} \ge 0

\\

\therefore \, &\left\vert e^{iz} \right\vert = e^{-R \sin(\theta)} =e^{\text{Re}(iz)}

\end{align}

Is it correct?

In these applications of the Residue Theorem, we often need to estimate the magnitude of the line integral of [itex] e^{iz} [/itex] over the semicircle [itex]= Re^{i\theta}, \; 0 \le \theta \le \pi[/itex]. We note that:

[tex]

\left\vert e^{iz} \right\vert = e^{\text{Re}(iz)} = e^{-R \sin(\theta)}

[/tex]

To understand the above, I did the following.

\begin{align}

&z = Re^{i\theta} = R \cos(\theta) + i R \sin(\theta)

\\

&iz = i R \cos(\theta) - R \sin(\theta)

\\

&e^{iz} = e^{i R \cos(\theta) - R \sin(\theta)} = e^{i R \cos(\theta)} e^{-R \sin(\theta)}

\\

&\left\vert e^{iz} \right\vert = \left\vert e^{i R \cos(\theta)} e^{-R \sin(\theta)} \right\vert

= \left\vert e^{i R \cos(\theta)} \right\vert \; \left\vert e^{-R \sin(\theta)} \right\vert

\\

&e^{i R \cos(\theta)} = e^{ix} && \text{where $x = R \cos(\theta)$, real}

\\

\therefore \, &\left\vert e^{i R \cos(\theta)} \right\vert = 1

&& \cos^2 x + \sin^2 x= 1

\\

&\left\vert e^{-R \sin(\theta)} \right\vert = e^{-R \sin(\theta)}

&& \text{since } e^{-R \sin(\theta)} \ge 0

\\

\therefore \, &\left\vert e^{iz} \right\vert = e^{-R \sin(\theta)} =e^{\text{Re}(iz)}

\end{align}

Is it correct?

Last edited: