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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.

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marked as duplicate by mweiss, Community Feb 22 '18 at 19:12

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  • $\begingroup$ Find the argument of the expression under the radical using what you know. Then halve it to find the argument of the square root. $\endgroup$ – Ethan Bolker Feb 22 '18 at 16:19
  • $\begingroup$ Thanks. Could you indicate any material that has this that you have suggested? $\endgroup$ – D.Silva Feb 22 '18 at 16:25
  • $\begingroup$ It's standard elementary complex arithmetic. Pick a site you like from this search for complex number arithmetic: google.com/… $\endgroup$ – Ethan Bolker Feb 22 '18 at 16:28
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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:

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

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

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

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  • $\begingroup$ Thank you. Is it correct to say that $arg(\lambda)$ is the principal argument? That is indeed what I am looking for. $\endgroup$ – D.Silva Feb 22 '18 at 17:07
  • $\begingroup$ Glad I could help. The answer is yes. And that is equal to $arg(z)/2$. @D.Silva $\endgroup$ – Mehrdad Zandigohar Feb 22 '18 at 17:20
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    $\begingroup$ Thanks for clarification. I did ContourPlot of $arg(\lambda)$ and I got what I expected. $\endgroup$ – D.Silva Feb 22 '18 at 17:46
  • $\begingroup$ @D.Silva Happy to hear that Silva. $\endgroup$ – Mehrdad Zandigohar Feb 22 '18 at 20:05
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make the ansatz $$\sqrt{x^2-y^2+m^2-2sixy}=A+Bi$$

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  • $\begingroup$ could you solve your Problem? $\endgroup$ – Dr. Sonnhard Graubner Feb 22 '18 at 17:10
  • $\begingroup$ Yes, In a slightly more laborious way. But I decided to follow the procedure of Sr. Merhdad Zandigohar. $\endgroup$ – D.Silva Feb 22 '18 at 17:13

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