Finding an argument of sum of two complex hyperbolic functions

Question

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?

Answer 1

$z_2$ 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 $$ \angle Q = {\text{arccot}} \frac{ -\cosh(x) \cos(y)+ a \sinh(x) \cos(y)\ - b \cosh(x) \sin(y) }{ -\sinh(x) \sin(y) + b \sinh(x) \cos(y) + a \cosh(x) \sin(y)} $$