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Complex Analysis : Complex Functions

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Example Questions

Example Question #1 : Complex Functions

What does the sum below equal?

$\sum_{n=1}^{m}(e^{2\pi i/m})^{n}$ $\displaystyle \sum_{n=1}^{m}(e^{2\pi i/m})^{n}$

Another way of asking this question is what is the sum of the $\displaystyle m$ roots of unity.

$m\in \mathbb{N}$ $\displaystyle m\in \mathbb{N}$

Possible Answers:

$\displaystyle 1$

$\pi$ $\displaystyle \pi$

$e^\pi$ $\displaystyle e^\pi$

$\displaystyle 0$

$\displaystyle e$

Correct answer:

$\displaystyle 0$

Explanation:

As messy as it looks, this is just a geometric series.

we will use the partial sum formula for the geometric series.

$S_{n}=\frac{a_{1}(1-r^n)}{1-r}$ $\displaystyle S_{n}=\frac{a_{1}(1-r^n)}{1-r}$

$a_{1}=e^{2\pi i/m}$ $\displaystyle a_{1}=e^{2\pi i/m}$

$r=e^{2\pi i/m}$ $\displaystyle r=e^{2\pi i/m}$

$S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i/m})^m})}{1-e^{2\pi i/m}}$ $\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i/m})^m})}{1-e^{2\pi i/m}}$

the red part is the only part that matters....the m's $\displaystyle m's$ cancel out leaving....

$S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i})})}{1-e^{2\pi i/m}}$ $\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (e^{2\pi i})})}{1-e^{2\pi i/m}}$

and...

$e^{2\pi i}=1$ $\displaystyle e^{2\pi i}=1$

thus we have...

$S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (1)})}{1-e^{2\pi i/m}}$ $\displaystyle S_{m}=\frac{a_{1}(1-r^m)}{1-r}=\frac{e^{2\pi i/m}(1-{\color{Red} (1)})}{1-e^{2\pi i/m}}$

which gives the answer of zero.

$S_{m}=0$ $\displaystyle S_{m}=0$

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Example Question #2 : Complex Functions

Consider the function $\displaystyle f(z) = |z|^2$

Find an expression for f'(z) $\displaystyle f'(z)$ (hint: use the definition of derivatve) and where it exists in the complex plane.

Possible Answers:

$f'(z) = 0 \text{ where } z=0$ $\displaystyle f'(z) = 0 \text{ where } z=0$

$f'(z) = 2|z| \text{ } \forall z$ $\displaystyle f'(z) = 2|z| \text{ } \forall z$

$f'(z) = 2z \text{ } \forall z$ $\displaystyle f'(z) = 2z \text{ } \forall z$

$f'(z) = 2z \text{ where } z=0$ $\displaystyle f'(z) = 2z \text{ where } z=0$

$f'(z) = 2|z| \text{ where } z=0$ $\displaystyle f'(z) = 2|z| \text{ where } z=0$

Correct answer:

$f'(z) = 0 \text{ where } z=0$ $\displaystyle f'(z) = 0 \text{ where } z=0$

Explanation:

Applying the definition of derivative, we have that

$f'(z) = \lim_{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} \\$ $\displaystyle f'(z) = \lim_{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} \\$

$f'(z) = \lim_{\Delta z \to 0}\frac{|z+\Delta z|^2 - |z|^2}{\Delta z} = \lim_{\Delta z \to 0} \frac{(z+\Delta z)(\bar{z}+\overline{\Delta z}) - z\overline{z}}{\Delta z} = \lim_{\Delta z \to 0} \overline{z} + \overline{\Delta z} + z \frac{\overline{\Delta z}}{\Delta z}$ $\displaystyle f'(z) = \lim_{\Delta z \to 0}\frac{|z+\Delta z|^2 - |z|^2}{\Delta z} = \lim_{\Delta z \to 0} \frac{(z+\Delta z)(\bar{z}+\overline{\Delta z}) - z\overline{z}}{\Delta z} = \lim_{\Delta z \to 0} \overline{z} + \overline{\Delta z} + z \frac{\overline{\Delta z}}{\Delta z}$

If the limits exists, it can be found by letting $\Delta z$ $\displaystyle \Delta z$ approach $\displaystyle 0$ in any manner.

In particular, if we it approach through the points $(\Delta x , 0 )$ $\displaystyle (\Delta x , 0 )$ , we have that

$\overline{\Delta z} = \overline{\Delta x + i0} = \Delta x - i0 = \Delta x + i0 = \Delta z$ $\displaystyle \overline{\Delta z} = \overline{\Delta x + i0} = \Delta x - i0 = \Delta x + i0 = \Delta z$

A similar approach with $(0, \Delta y)$ $\displaystyle (0, \Delta y)$ implies that $\overline{\Delta z } = -\Delta z$ $\displaystyle \overline{\Delta z } = -\Delta z$

Since limits are unique, these two approaches imply that

$\overline{z} + z = \overline{z} - z$ $\displaystyle \overline{z} + z = \overline{z} - z$ , which implies z=0 $\displaystyle z=0$ and f'(z) $\displaystyle f'(z)$ cannot exist when $z \neq 0$ $\displaystyle z \neq 0$

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Complex Analysis : Complex Functions

Study concepts, example questions & explanations for Complex Analysis

All Complex Analysis Resources

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Example Question #1 : Complex Functions

Example Question #2 : Complex Functions

All Complex Analysis Resources

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