Can Wolfram Alpha Handle Complex Inequality Solutions?

[╔(σ_σ)╝]

lol sorry and thanks for your help!

What did you do here ?

( SQRT( 1 + #^2)SQRT( 1 + #^2) ) / ( (1 + #i)SQRT( 1 + #^2) )
(1 + #^2)/( (1 + #i)SQRT( 1 + #^2) )
(1 - ix)/SQRT( 1 + #^2)- >>>>>>>>>>>>>>>>>>>>>>>>>>>>>?
1/SQRT( 1 + #^2 ) - (ix)/SQRT( 1 + #^2)

How did you go from step 2 to 3 ?

 

[jackmell]

Homework Statement

I was trying rewrite properly the tangent and arctangent function and was trying to figure out what values of the argument in the natural log was < 0 so that way I could simplify my formulas further because

i ln(-1) = -pi/2
and stuff of the sort

(-x i)/SQRT(x^2 + 1) + SQRT(x^2 +1)/(x^2 + 1) < 0
basically looking for the first x value that would make that a negative number

I'd say x=0 is the only answer that makes the expression a negative number (zero imaginary part) assuming you don't use the principal square root values. That is, use the function:

$$f(x)=-\frac{x i}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{x^2 + 1}$$

or:

$$f(x)=\frac{x i}{\sqrt{x^2 + 1}} - \frac{\sqrt{x^2 + 1}}{x^2 + 1}$$

where the root symbol means "principal value".

Also, Wolfram Alpha and Mathematica only use the principal value by default when evaluating a multi-valued function.