In my class, my instructor told us that the square root of a complex number is in general not a function because it is multi-valued. For example, e^(ipi/4) could have a square root of e^(ipi/8) or e^(i*9pi/8). He then added that if we shortened the domain of the polar angle to (-pi,pi), the square root then becomes a function. I don't see how this works. The square root has a period of pi, over which it repeats itself. For eq, both e^(-ipi/4) and e^(ipi/4) qualify as square roots of e^(i*pi/2), and they both lie in the shortened domain. I don't see where I am going wrong.
Subscribe to:
Post Comments (Atom)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
How can one proof the equality $$\sum\limits_{v=0}^k \frac{k^v}{v!}=\sum\limits_{v=0}^k \frac{v^v (k-v)^{k-v}}{v!(k-v)!}$$ for $k\in\mathbb{...
-
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
If you think about the limit definition of the derivative, $dy$ represents $$\lim_{h\rightarrow 0}\dfrac {f(x+h)-f(x)}{h}$$, and $dx...
No comments:
Post a Comment