Square root argument of a complex number.

Question

How could I compute the argument of the following complex number: $$\lambda=\sqrt{x^2-y^2+m^2-2isxy}$$ where, $s=\pm1,\,\,m\in\mathbb{R}$. I know for number in algebraic form $z=a+bi$ it's easy, just do, $arg(z)=\arctan\left(\dfrac{b}{a}\right)$ but unfortunately this is not the case. Thank you in advanced.

Answer 1

• Mostly when you are facing radicals, converting to polar coordinates is the key:

$z={x^2-y^2+m^2-2isxy}=a+bi$

$arg(z)=\arctan\left(\dfrac{b}{a}\right)$

$\|z\|=\sqrt{a^2+b^2}$

$\displaystyle\lambda=\|z\|^{1/2}e^{i\times arg(z)/2}$

• There is also another approach but I could not continue it:

$\displaystyle \lambda^2 =(x\pm iy)^2+m^2$

$\lambda_1=\sqrt{z^2+m^2}$

$\lambda_2=\sqrt{\bar z ^2+m^2}$

Answer 2

make the ansatz $$\sqrt{x^2-y^2+m^2-2sixy}=A+Bi$$