COMPLEX ANALYSIS Finding the inverse of the function g(z) = Log(z-i) and the domain.

I have a problem with showing that this complex function is one-one. Could someone help me please?

Show that the function g(z) has an inverse, and find its domain and rule.

g(z) = Log ( z-i )

I know the C - { i } is the domain of g, but I am unsure of the codomain. I know g^-1 will have codomain C-{i}.

I have let w = g(z) so that the rule for g^-1 = e^w + i which I am happy with. My problem is showing now that g^-1 is one-one, so that the inverse exists, and finding the codomain of g.

I have a problem with showing that this complex function is one-one. Could someone help me please?

Show that the function g(z) has an inverse, and find its domain and rule.

g(z) = Log ( z-i )

I know the C - { i } is the domain of g, but I am unsure of the codomain. I know g^-1 will have codomain C-{i}.

I have let w = g(z) so that the rule for g^-1 = e^w + i which I am happy with. My problem is showing now that g^-1 is one-one, so that the inverse exists, and finding the codomain of g.

Hi Log is the complex logarithm defined as:

For z in the set C - {0} the principal logarithm of z is

Log z = ln|z| + i Arg|z|

It is the inverse of the function f(z) = e^z such that z is in the set {x+iy:-pi<y<=pi}

I have got this so far

Let g(z) = Log(z - i) so the domain of g is C - {i}. Now let C - {i}= A, then

g(A) = {Log(z-i) z: z is in the set C - {i} }

g(A) = {w=Log(z-i) z: z is in the set C - {i} }

g(A) = {w= ln|z-i| + i Arg(z-i) z: z is in the set C - {i} }

Now this is where I can't SEE the answer to what the codomain is. I think that w needs to be in the set C minus anything that makes the argument of w zero, but this could be incorrect!

..

I'm not familiar with the term codomain (and the wiki page defn doesn't seem to me like it's what you're after(*)), but as RichE says, the range of values taken by Log(z-i) is going to be identical to the range taken by Log(z), so I would expect Log(z-i) to have the same codomain as Log(z).

(*) As defined in the wiiki, it's just the destination set when you say "f(z) is a function from X to Y". Since in this context it seems there are various possible choices for Y (some larger than others) it didn't really seem to fit what you're after.

Thanks for your replies both. That makes it clearer now, but I'm not sure what the range of Log(z) is? This is where I keep going round in circles. Is it that the principal Logz has range -pi<Im(w)<=pi, and then I can say that Log(z-i) will have the same range?

Yes. (Because for any "input" z you put into Log z, if you choose w = z+i as the "input" to Log(w-i), you're obviosuly going to get the same result.

It is all very clear now. I really appreciate your help. Thank you!

Let g(z) = Log(z - i) so the domain of g is C - {i}. Now let C - {i}= A, then

g(A) = {Log(z-i) z: z is in the set C - {i} }

g(A) = {w=Log(z-i) z: z is in the set C - {i} }

g(A) = {w= ln|z-i| + i Arg(z-i) z: z is in the set C - {i} }

Now this is where I can't SEE the answer to what the codomain is. I think that w needs to be in the set C minus anything that makes the argument of w zero, but this could be incorrect!

It seems to me that you could formalise this a bit as follows:

Let $w=\ln |z-i|+i \rm{Arg}(z-i)$ with $z \in \mathbb{C}-\{i\}$

Let $u=z-i$ then

$u+i=z \in \mathbb{C}-\{i\} \Rightarrow u+i \ne i \Rightarrow u \ne 0 \Rightarrow u \in \mathbb{C}-\{0\}$

so $w=\ln |u|+i \rm{Arg}(u)$ with $u \in \mathbb{C}-\{0\}$

Or have I missed the point?

I feel this misses the point, yes. Range of Log z is given and obviously range of Log(z-i) is going to be the same.