What is the Taylor expansion for ln(1+z)?

[Axiom_137]

For a common ratio |z| < 1, we have:
$$\frac{a}{1 - z} = \sum_{k = 0} ^ {\infty} az ^ k$$, so apply it here, we have:
$$\frac{1}{1 + z} = \sum_{k = 0} ^ {\infty} (-z) ^ k, \ \forall |z| < 1$$ (a geometric series with a common ratio -z, and the first term is 1).
Integrate both sides gives:
$$\int \frac{dz}{1 + z} = \int \left( \sum_{k = 0} ^ {\infty} (-z) ^ k \right) dz$$

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

$$\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz$$

$$ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}$$

Plug in z=0 and the sum is zero...My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

 

[d_leet]

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

$$\left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz$$

$$ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}$$

Plug in z=0 and the sum is zero...


My calculator says ln(i) is not zero. ln(i)= i*(pi/2)

So is there seems to be a problem using this method with complex variables.

(Or have I made a mistake with my math?)

I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.

 

[Axiom_137]

I think that sum only holds for |z|<1, that is strictly less than one, whereas |i|=1 so one should not expect the equation to hold for z=i since i does not satisfy the necessary inequality.

For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1

 

[d_leet]

For the geometric series, the inequality for this problem would be :

|-iz|<1 or simply |z|<1

z=0 satisfies |z|<1

Yes you're absolutely right, I have no idea what I was thinking, and I cannot think of a reason as to why the sum does not work other than the fact that the series does, at least, correctly calculate the real part of ln(i)

 

[HallsofIvy]

Please, tell us what you think the derivative of 1/(1+ z) is!

 

[Hurkyl]

This seems a little too simple for the complex plane.

What about ln(i-z)? hmmm...

$$\int \left( \frac{1}{i} \right)\frac{dz}{1 + iz} = \int \left( -i\sum_{k = 0} ^ {\infty} (-iz) ^ k \right) dz$$

$$ln(i-z) = \sum_{k = 0} ^ {\infty} \frac{(-iz)^{k+1}}{(k+1)}$$

Plug in z=0 and the sum is zero...

You forgot the constant of integration.