I want to find the absolute value and the argument of the following complex number $$ -\cosh[\sqrt{z_1}]+\frac{z_2}{\sqrt{z_1}}\sinh[\sqrt{z_1}] $$ where $z_1$ and $z_2$ are two complex numbers. Can anyone help?

  • $\begingroup$ It is true that you did not show any effort, but I think that the downvote is a bit unfair, the question is interesting. To improve it, you can at least specify what do you mean by the square root. As you know, each complex number has two distinct square roots, which one do you choose? $\endgroup$ – Giuseppe Negro Oct 16 '18 at 14:49
  • $\begingroup$ hint: math.stackexchange.com/questions/44406 $\endgroup$ – Andreas Oct 16 '18 at 15:42

$z_2$ appears only once. So let $\frac{z_2}{\sqrt{z_1}} = a + i b$. Further, let $\sqrt{z_1} = x + iy$. The $a,b,x,y$ are real.

Then you have $$ Q = -\cosh[\sqrt{z_1}]+\frac{z_2}{\sqrt{z_1}}\sinh[\sqrt{z_1}] = -\cosh[x + iy]+(a+ib)\sinh[x + iy] $$

Use the two identites (see here) $$ \cosh(x+iy) = \cosh(x) \cos(y) + i \sinh(x) \sin(y) \\ \sinh(x+iy) = \sinh(x) \cos(y) + i \cosh(x) \sin(y) $$ to obtain $$ Q = -\cosh(x) \cos(y) - i \sinh(x) \sin(y)+(a+ib)(\sinh(x) \cos(y) + i \cosh(x) \sin(y) )\\ = -\cosh(x) \cos(y)+ a \sinh(x) \cos(y)\ - b \cosh(x) \sin(y) + \\ + i \left[-\sinh(x) \sin(y) + b \sinh(x) \cos(y) + a \cosh(x) \sin(y)\right] $$

So we have $$ |Q|^2 = \left[-\cosh(x) \cos(y)+ a \sinh(x) \cos(y)\ - b \cosh(x) \sin(y) \right]^2+ \\ + \left[-\sinh(x) \sin(y) + b \sinh(x) \cos(y) + a \cosh(x) \sin(y)\right]^2 $$

and $$ \angle Q = {\text{arccot}} \frac{ -\cosh(x) \cos(y)+ a \sinh(x) \cos(y)\ - b \cosh(x) \sin(y) }{ -\sinh(x) \sin(y) + b \sinh(x) \cos(y) + a \cosh(x) \sin(y)} $$

  • $\begingroup$ seems fine but how would you write your final result in terms of $z_1$ and $z_2$. I mean what is the a,b,x and y in terms of $z_1$ and $z_2$. $\endgroup$ – Parveen Oct 16 '18 at 11:25
  • $\begingroup$ Well that should be easy. By the way: have a look at math.meta.stackexchange.com/questions/9959 which should make it clear that you have to provide some effort by yourself - so far you have demonstrated nothing. $\endgroup$ – Andreas Oct 16 '18 at 11:29

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