http://www.math.ualberta.ca/~runde/files/ass411-1.pdf
I can't do #3 or #4
I have a feeling I can do 3 but I'm not quite sure where to go from
and for 4 I know that to show it's not injective show that exp(0,0) = exp(0,2pi)
http://www.math.ualberta.ca/~runde/files/ass411-1.pdf
I can't do #3 or #4
I have a feeling I can do 3 but I'm not quite sure where to go from
and for 4 I know that to show it's not injective show that exp(0,0) = exp(0,2pi)
3)The problem here is that how is really defined what $\displaystyle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ means? I assume it means that we think of $\displaystyle f: D\to \mathbb{C}$ as $\displaystyle f(x,y) = (u(x,y),v(x,y))$ (since the complex numbers are simply points in $\displaystyle \mathbb{R}^2$ formally).
Now define $\displaystyle \frac{\partial f}{\partial x} = \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right)$ and $\displaystyle \frac{\partial f}{\partial y} = \left( \frac{\partial u}{\partial y}, \frac{\partial v}{\partial y} \right)$.
This means $\displaystyle \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right) + i \left( \frac{\partial u}{\partial y}, \frac{\partial v}{\partial y} \right) = \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right) + \left( - \frac{\partial v}{\partial y} , \frac{\partial u}{\partial y} \right)$.
However, $\displaystyle \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 0$ and $\displaystyle \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} = 0$ by Cauchy-Riemann equations.
4)Note $\displaystyle \exp(0) = \exp(2\pi i)$ thus it is not injective.
May I offer a different perspective?
If $\displaystyle x=\frac{z+\overline{z}}{2}$ and $\displaystyle y=\frac{z-\overline{z}}{2i}$ then:
$\displaystyle
\frac{\partial f}{\partial \overline{z}}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \overline{z}}+i\left(\frac{\partial v}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial \overline{z}}\right)$
.....$\displaystyle =\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
$
Well there you go: If $\displaystyle f$ is analytic then the partials satisfy the Cauchy-Riemann equations so $\displaystyle \frac{\partial f}{\partial \overline{z}}=0$ and the only way for
$\displaystyle \frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)=0$
is for the partials to satisfy the Cauchy-Riemann equations, that is, $\displaystyle f$ must be analytic.
thank you but i got that one already
also for #4
where it says show that exp(C) = C/{0}
what does that mean? the notation is unfamiliar to me and hasn't been explained in course lecture or notes
also there is no textbook for the class which makes this particularly frustrating
Hey jb. It's a function at a domain. Suppose I define the unit circle as the domain $\displaystyle D=\left\{z : |z|<=1\right\}$
Then I could say, what is the function at that domain? or what is $\displaystyle f(D)$ meaning how does the function map the domain $\displaystyle D$. Same with $\displaystyle e^{\mathbb{C}}$. How does the exponential function map all of $\displaystyle \mathbb{C}$? It does so by the expression:
$\displaystyle e^{\mathbb{C}}=\mathbb{C}\backslash\{0\}$ meaning it maps all of $\displaystyle \mathbb{C}$ to the deleted neighborhood:$\displaystyle \mathbb{C}\backslash\{0\}$ which is $\displaystyle \mathbb{C}$ minus the origin.
If $\displaystyle \alpha \in \mathbb{C}^{\times}$ then you should know the fact that is is possible to write $\displaystyle \alpha = re^{i\theta}$ where $\displaystyle r>0$. Now let $\displaystyle z = \log r + i\theta$. Then $\displaystyle e^z = e^{\log r + i \theta} = re^{i\theta} = \alpha$. This shows that the equation $\displaystyle e^z = \alpha$ always has a solution for any $\displaystyle \alpha\in \mathbb{C}^{\times}$. Thus, the function $\displaystyle \exp : \mathbb{C} \to \mathbb{C}^{\times}$ is onto.
It may be late (since it's past 3pm...), but I don't care
I'm also learning how to do this stuff !
For #5 c), use Cauchy-Hadamard theorem : Cauchy-Hadamard theorem - Wikipedia, the free encyclopedia
My teacher says the best way above all to solve this problem is to use Abel's theorem :
$\displaystyle R=\sup \{ ~ r \in \mathbb{R} / a_n r^n \text{ is bounded } \}$(R is the radius of convergence)
Here, Hadamard's theorem is more straightforward.
#6 is above my capacities...