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- Thread starter dwsmith
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- Feb 13, 2012

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May be it has more sense the identity...Find the real functions \(u\) and \(v\) such that \(f = u + iv\) for \(f(z) = \arctan(z)\).

Does below work or make sense to do?

\[

\arctan(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}

\]

$\displaystyle \tan^{-1} z = \frac{i}{2}\ \ln \frac{1 - i\ z}{1 + i\ z}\ (1)$

Now set in (1) $z = x + i\ y$ and separe the real and imaginary parts...

Kind regards

$\chi$ $\sigma$

- Jan 29, 2012

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[tex]\arctan(z)= \frac{1}{z}[/tex] certainly does NOT make sense- it is not true. What made you write that?Find the real functions \(u\) and \(v\) such that \(f = u + iv\) for \(f(z) = \arctan(z)\).

Does below work or make sense to do?

\[

\arctan(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}

\]

I'd consider writing [tex]\displaystyle \begin{align*} z = x + i\,y \end{align*}[/tex], then [tex]\displaystyle \begin{align*} f(z) = u(x,y) + i\,v(x,y) \end{align*}[/tex], thenFind the real functions \(u\) and \(v\) such that \(f = u + iv\) for \(f(z) = \arctan(z)\).

Does below work or make sense to do?

\[

\arctan(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}

\]

[tex]\displaystyle \begin{align*} \arctan{ (x + i\,y) } &= u + i\,v \\ x + i\,y &= \tan{ \left( u + i\,v \right) } \end{align*}[/tex]

and trying to manipulate the tangent function so you can write it in terms of its real and imaginary parts...

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In doing so, we haveMay be it has more sense the identity...

$\displaystyle \tan^{-1} z = \frac{i}{2}\ \ln \frac{1 - i\ z}{1 + i\ z}\ (1)$

Now set in (1) $z = x + i\ y$ and separe the real and imaginary parts...

Kind regards

$\chi$ $\sigma$

\[

\frac{i}{2}\ln(1 + y - ix) - \frac{i}{2}\ln(1 - y + ix)

\]

Let \(z_1 = 1 + y - ix\) and \(z_2 = 1 - y + ix\).

Then \(\lvert z_1\rvert = \sqrt{(1 + y)^2 + x^2}\), \(\varphi_1 = -\arctan\left(\frac{x}{1 + y}\right)\), \(\lvert z_2\rvert = \sqrt{(1 - y)^2 + x^2}\), and \(\varphi_2 = -\arctan\left(\frac{x}{1 - y}\right)\).

Mathematica has that one term of the real part is

\[

\frac{1}{4}\text{Im}\left(\ln\left[\frac{(1 + y)^2 + x^2}{(1 - y)^2 + x^2}\right]\right)

\]

but when is the argument of the ln imaginary? Is it ever?

Last edited:

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\(\DeclareMathOperator{\Arg}{Arg}\)

How do I go from

\[\frac{i}{4}\ln\Bigg[\frac{(1 + y)^2 + x^2}{(1 - y)^2 + x^2}\Bigg] -

\frac{1}{2}\Arg\bigg[\frac{1 + y - ix}{1 - y + ix}\bigg]

\]

to the real part is

\[

\frac{1}{2} \left(-\arg \left(\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{x^2+(y-1)^2}}\right)-\arg \left(-1-\frac{2}{-i x+y-1}\right)\right)

\]

More specifically, how is \(-\arg \left(\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{x^2+(y-1)^2}}\right)\) obtained?

How do I go from

\[\frac{i}{4}\ln\Bigg[\frac{(1 + y)^2 + x^2}{(1 - y)^2 + x^2}\Bigg] -

\frac{1}{2}\Arg\bigg[\frac{1 + y - ix}{1 - y + ix}\bigg]

\]

to the real part is

\[

\frac{1}{2} \left(-\arg \left(\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{x^2+(y-1)^2}}\right)-\arg \left(-1-\frac{2}{-i x+y-1}\right)\right)

\]

More specifically, how is \(-\arg \left(\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{x^2+(y-1)^2}}\right)\) obtained?

Last edited:

- Jan 29, 2012

- 1,151

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No.

I want to find the real and imaginary parts of \(\arctan(z)\). I came up with the \(\frac{1}{z}\) by using a right triangle; however, that was noted to be incorrect. I have the form of \(\arctan(z)\) but I don't know how Mathematica achieved part of its real part solution which I asked in my next most recent post.