# Complex real function u when f = u + iv

#### dwsmith

##### Well-known member
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?
$\arctan(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}$

#### chisigma

##### Well-known member
Re: complex real function u when f = u + iv

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

$\chi$ $\sigma$

#### HallsofIvy

##### Well-known member
MHB Math Helper
Re: complex real function u when f = u + iv

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?
$\arctan(z) = \frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2}$
$$\arctan(z)= \frac{1}{z}$$ certainly does NOT make sense- it is not true. What made you write that?

#### Prove It

##### Well-known member
MHB Math Helper
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 \displaystyle \begin{align*} z = x + i\,y \end{align*}, then \displaystyle \begin{align*} f(z) = u(x,y) + i\,v(x,y) \end{align*}, then

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

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

#### dwsmith

##### Well-known member
Re: complex real function u when f = u + iv

May 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

$\chi$ $\sigma$
In doing so, we have
$\frac{i}{2}\ln(1 + y - ix) - \frac{i}{2}\ln(1 - y + ix)$
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?

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#### dwsmith

##### Well-known member
$$\DeclareMathOperator{\Arg}{Arg}$$
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?

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#### HallsofIvy

##### Well-known member
MHB Math Helper
You titled this "complex function". But it appears you are trying to solve the equation arctan(z)= 1/z. Is that correct?

#### dwsmith

##### Well-known member
You titled this "complex function". But it appears you are trying to solve the equation arctan(z)= 1/z. Is that correct?
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.