Complex Integral: Solving from Ln to ArcTan

[sam2]

Hi,

I'm doing the following as an exercise to try and get my head around complex numbers. Specifically, I need to understand what it means to take the natural log of a complex number and what it involves.

Say I wanted to integrate 1/ (1 +x^2) dx

I know this is arcTan(x).


I can also write the integral as dx / [ (1 + ix)(1 - ix) ], which can be solved with partial fractions to arrive at 0.5 Ln [(1 + ix) / (1 - ix) ].

Now it should follow that the 2 results are equivalent. How do I got from Ln to arctan? Any hints?

Mant Thanks for your help,

 

[amcavoy]

Hi,
I'm doing the following as an exercise to try and get my head around complex numbers. Specifically, I need to understand what it means to take the natural log of a complex number and what it involves.
Say I wanted to integrate 1/ (1 +x^2) dx
I know this is arcTan(x).
I can also write the integral as dx / [ (1 + ix)(1 - ix) ], which can be solved with partial fractions to arrive at 0.5 Ln [(1 + ix) / (1 - ix) ].
Now it should follow that the 2 results are equivalent. How do I got from Ln to arctan? Any hints?
Mant Thanks for your help,

$$\tan{x}=\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$$

 

[Lonewolf]

Looks like you missed a factor of -i of. When you integrate something like $\frac{1}{1+ix}$, you will get something like the result $-iLn(1+ix)$. Now go and compare the Taylor Series of arctan(x) and your result

 

[sam2]

Gotcha.

Many thanks.

 

[George Jones]

How do I got from Ln to arctan? Any hints?

Sorry, LateX preview is not working for me.

Consider a complex number z = a + ib. In polar form, z = re^{i theta} with r = sqrt{a^2 + b^2} and theta = tan^(-1){b/a}. Now take |z| = r = 1, so b = sqrt{1 - a^2}. Take the ln of both sides of e{^i theta} = a + i sqrt{1 - a^2}.

i theta = ln(a + i sqrt{1 - a^2})

i tan^(-1){sqrt{1 - a^2}/a} = ln(a + i sqrt{1 - a^2})

i tan^(-1){x} = ln{a(1 + ix)}, where x = sqrt{1 - a^2}/a

Now invert the x equation to find a and you're home.

Regards,
George

 

[sam2]

Hi George,

Thanks for the reply but I am still stuck when I use your method. It makes perfect sense but I can't see where I am going wrong. I am doing the integral that I mentioned in my first post; but basically end up with the integral equal to tan^{-1} [ -2x/(1-x^2) ] Sorry, latex doesn't seem to be working for me!

A brief outline:

I re-write the integral as

dx / (1-ix)(1+ix)

Using partial fractions this comes up to:

0.5/(1-ix) dx + 0.5/(1+ix) dx

I do the integral,

0.5.i. Ln[(1-ix) / (1+ix)]

then write everything as r.e^{i.theta} . r turns out to be 1 as your mentioned in yoru post, but I end up with something like

-0.5 arctan{ -2x/(1-x^2) }

Can you sugegst where I might be going wrong?

Thanks,

 

[George Jones]

i tan^(-1){x} = ln{a(1 + ix)}, where x = sqrt{1 - a^2}/a

Squaring the second equation gives

a^2 x^2 = 1 - a^2

a^2 (1 + x^2) = 1

a^2 (1 +ix)(1 -ix) = 1

a = [(1 +ix)(1 -ix)](-1/2)

Using this in the first equation above gives

i tan^(-1){x} = ln{[(1 +ix)(1 -ix)](-1/2) (1 + ix)}

i tan^(-1){x} = ln{[(1 + ix)/(1 - ix)]^(1/2)}

i tan^(-1){x} = 1/2 ln{(1 + ix)/(1 - ix)}

tan^(-1)(x} = -i/2 ln{(1 + ix)/(1 - ix)}

tan^(-1)(x} = i/2 ln{(1 - ix)/(1 + ix)}

This is the same result as yours, but achieved without integration, i.e., the 2 different methods give consistent results.

Regards,
George

 

[sam2]

much appreciated