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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.

D.Silva
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  • Find the argument of the expression under the radical using what you know. Then halve it to find the argument of the square root. – Ethan Bolker Feb 22 '18 at 16:19
  • Thanks. Could you indicate any material that has this that you have suggested? – D.Silva Feb 22 '18 at 16:25
  • It's standard elementary complex arithmetic. Pick a site you like from this search for complex number arithmetic: https://www.google.com/search?q=complex+number+arithmetic&ie=utf-8&oe=utf-8&client=firefox-b-1 – Ethan Bolker Feb 22 '18 at 16:28

2 Answers2

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

Dr. Sonnhard Graubner
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  • 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:

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

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

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

Mehrdad Zandigohar
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