Magnitude of a complex expression

Question

Is there a way to derive an expression for the magnitude of $$ \frac{2 + (1-2ia\lambda \sin \theta)^{1/2}}{3 + 2ia\lambda\sin\theta} $$

I know how to do this if the square root weren't there. Any hint would be appreciated.

Answer 1

To determine the modulus, we need to multiple by the conjugate which is $3 - 2ia\lambda\sin(\theta)$. Now, let $w = (1 - 2ia\lambda\sin(\theta))^{1/2}$ which has a modulus of $\lvert w\rvert = \sqrt{1 + 4a^2\lambda^2\sin^2(\theta)}$ and the principle argument is $\phi = \arctan(-2a\lambda\sin(\theta))$ where $\phi\in(-\pi, \pi)$. Therefore, we can write $$ (1 - 2ia\lambda\sin(\theta))^{1/2} = \exp(1/2(\ln\lvert w\rvert + i\phi)) = w^{1/4}e^{i\phi/2} $$ Now the denominator is a real number and the numerator no longer has a complex number in a square root. All you need to do is separate the real and imaginary parts so you can find the modulus of $z$ $$ \frac{(2 + \lvert w\rvert^{1/2}e^{i\phi/2})(3 - 2ai\lambda\sin(\theta))}{9+4a^2\lambda^2\sin^2(\theta)} $$