Find $i\log(x-i)+i^2\pi+i^3\log(x+i)+i^4(2\arctan x)$, if $x>0$

Question

Find $i\log(x-i)+i^2\pi+i^3\log(x+i)+i^4(2\arctan x)$, if $x>0$

The equation can be written as $$y=i\log(x-i)-\pi-i\log(x+i)+2\arctan x$$ $$y=i\log\frac{x-i}{x+i}-\pi+2\arctan x$$ Let $x+i=re^{i\theta}$ so $x-i=re^{-i\theta}$. $$y=i\log(e^{-2i\theta})-\pi+2\arctan x$$ $\theta=\arctan\frac{1}{x}=\cot^{-1}x$ $$y=2\cot^{-1}x-\pi+2\tan^{-1}x=2\cdot\frac{\pi}{2}-\pi=0$$

Is this correct?

Maybe this relation between the complex logarithm and the arctangent is useful here: http://math.stackexchange.com/questions/1102389/arctan-in-complex-logarithm-form — Dmoreno, Dec 26 '15 at 14:58
I think you should first rationalize $\frac{x-i}{x+i}$ and then use $\text{Log} z=\ln|z|+i\text{Arg} z$ — Aditya Agarwal, Dec 26 '15 at 15:01
@AdityaDev There is a difference between logarithm of real numbers and logarithm of complex numbers. Do you know that? — SchrodingersCat, Dec 26 '15 at 15:01
Consider the polynomial in $a$: $a\log(x-a)+a^2\pi+a^3\log(x+a)+a^4(2\arctan a)$. — Aditya Agarwal, Dec 26 '15 at 15:06
Why should I make the problem more difficult @Aditya Agarwal ? — Aditya Dev, Dec 26 '15 at 15:33

score 3 · Accepted Answer · answered Dec 26 '15 at 15:06

The last step is wrong.

Please note the following:

If $x+i=re^{i\theta}$, then $\theta=$ arg (x+i) = $\arctan \frac{1}{x}=arccot x$

So,

$$i\log(e^{-2i\theta})-\pi+2\arctan x$$ $$=i(\log|1|+(-2i\theta))-2(\frac{\pi}{2}-\arctan x)$$ $$=i(\log|1|-2i\theta)-2arccot x$$ $$=2arccot x-2arccot x$$ $$=0$$

Hope this helps.

Jan Eerland · Answer 2 · 2015-12-26T18:53:21.253

Assume $x\in\mathbb{R^+}$:

$$y(x)=i\ln(x-i)+i^2\pi+i^3\ln(x+i)+i^4\left(2\arctan(x)\right)=$$ $$i\ln(x-i)-\pi-i\ln(x+i)+1\left(2\arctan(x)\right)=$$ $$i\ln(x-i)-\pi-i\ln(x+i)+2\arctan(x)=$$ $$2\arctan(x)-\pi+i\left(\ln(x-i)-\ln(x+i)\right)=$$ $$2\arctan(x)-\pi+i\left(\ln\left(\frac{x-i}{x+i}\right)\right)=$$ $$2\arctan(x)-\pi+i\left(-2i\text{arccot}(x)\right)=$$ $$2\arctan(x)-\pi+2\text{arccot}(x)=$$ $$2\left(\arctan(x)+\text{arccot}(x)\right)-\pi=$$ $$2\left(\arctan(x)+\arctan\left(\frac{1}{x}\right)\right)-\pi=2\left(\frac{\pi}{2}\right)-\pi=\pi-\pi=0$$

So:

$$y(x)=2\left(\arctan(x)+\text{arccot}(x)\right)-\pi=0\space\space\space\space\space\space\space\space\space\space\text{with}\space\space x\in\mathbb{R^+}$$

Find $i\log(x-i)+i^2\pi+i^3\log(x+i)+i^4(2\arctan x)$, if $x>0$

2 Answers2