eq (2) pg 513 well defined for non-injective f ?

mark

I would also suggest that you read chapter 5 VI on pages 229 to 231.

Vasco/Admin

Mark

Since definition (2) on page 513 is written with an identically equal sign

$\widetilde{\Phi}[f(z)]\equiv\Phi(z)$

we should write

$\widetilde{\Phi}(1) = \widetilde{\Phi}[ f(1) ]\equiv \Phi(1)$
$\widetilde{\Phi}(1) = \widetilde{\Phi}[ f(e^{i2\pi/3}) ]\equiv \Phi(e^{i2\pi/3})\equiv\Phi(1)$
$\widetilde{\Phi}(1) = \widetilde{\Phi}[ f(e^{-i2\pi/3}) ] \equiv \Phi(e^{-i2\pi/3})\equiv\Phi(1)$

So the values of $\Phi(1),\Phi(e^{i2\pi/3}),\Phi(e^{-i2\pi/3})$ are all defined to be identically equal to $\Phi(1)$.

So $\Phi$ and $\widetilde{\Phi}$ cannot be multifunctions by definition (i.e. definition (2) on page 513).

Vasco/Admin

Mark

Equation (2) on page 513 means that corresponding points in the $z$-plane and $w$-plane are assigned equal function values.
If, as in your first example, we have $w=f(z1)=f(z2)=f(z3)$, then our definition in (2) means that

$\widetilde{\Phi}(w1)=\Phi(z1)$
$\widetilde{\Phi}(w2)=\Phi(z2)$
$\widetilde{\Phi}(w3)=\Phi(z3)$

But if $w1=w2=w3=w$ then from our definition (2) we must assign a single value to $\widetilde{\Phi}(w)$ and since
$\Phi(z1)=\widetilde{\Phi}(w)$ and $\Phi(z2)=\widetilde{\Phi}(w)$ and $\Phi(z3)=\widetilde{\Phi}(w)$ then we must assign the same value of $\Phi$ to all 3 points $z1,z2,z3$ and it must be the value $\widetilde{\Phi}(w)$.

Don't forget we are in charge and we are defining $\widetilde{\Phi}$ and $\Phi$ according to (2). Note the use of the word 'assign' and not calculate. We want $\widetilde{\Phi}$ and $\Phi$ to be the same for corresponding points in the $z$ and $w$ planes.

So we could write
$\widetilde{\Phi}(w=1)=100$ and $\Phi(z=1)=100$ and $\Phi(z=e^{2\pi/3})=100$ and $\Phi(z=e^{-2\pi/3})=100$
But of course 100 could be replaced by a more complicated function.

Vasco

Vasco,

why is $f=\sqrt[3]{z}$ multivalued while $f = z^3$ is not?

Thank you.