What is the absolute value of the following?

Question

So the expression is as follows

$$ \frac{e^{-jk \left(r+d\cos\frac{\theta}{2}\right)}}{r} + \frac{e^{j\varphi}e^{-jk \left(r-d\cos\frac{\theta}{2}\right)}}{r} $$

I factored out the $\frac{e^{-jkr}}{r}$ and took the absolute value of this which is $\frac{1}{r}$ this leaves the absolute value of $e^{-jkd\cos\theta/2} + e^{j\phi}e^{jkd\cos\theta/2}$.

However, I can't solve the remaining term and hoped for help.

Answer 1

Note

$$|e^{-jkd\cos\theta/2} + e^{j\phi}e^{jkd\cos\theta/2}|^2$$ $$=(e^{-jkd\cos\theta/2} + e^{j\phi}e^{jkd\cos\theta/2}) (e^{jkd\cos\theta/2} + e^{-j\phi}e^{-jkd\cos\theta/2})$$ $$=1+ e^{j\phi}e^{2jkd\cos\theta/2} + e^{-j\phi}e^{-2jkd\cos\theta/2} +1$$ $$=2+ 2\cos(\phi+2kd\cos\theta/2) $$

Thus, the absolute value is

$$\frac{\sqrt2}r\sqrt{1+\cos(\phi+2kd\cos\theta/2) }$$

Answer 2

For notational simplicity, let $a=r+d\cos\frac{\theta}{2},\ b=r-d\cos\frac{\theta}{2}$. Recall that the magnitude of a complex number, $|z|=\sqrt{z\cdot z^*}$, is the square root of the product of the number with its conjugate and that for all complex numbers, $(z+w)^*=z^*+w^*$ and $\left(\frac{z}{w}\right)^*=\frac{z^*}{w^*}$, and that the conjugate of a number in polar form is $re^{i\theta}\mapsto re^{-i\theta} $. Then, the square of the magnitude is $$|z|^2=\left(\frac{e^{-jka}}{r}+\frac{e^{j\varphi}e^{-jkb}}{r}\right)\left(\frac{e^{-jka}}{r}+\frac{e^{j\varphi}e^{-jkb}}{r}\right)^{*} \\ =\frac{\left(e^{-jka}+e^{j\left(\varphi-kb\right)}\right)}{r}\frac{\left(\left(e^{-jka}\right)^{*}+\left(e^{j\left(\varphi-kb\right)}\right)^{*}\right)}{r^{*}} \\ =\frac{\left(e^{-jka}+e^{j\left(\varphi-kb\right)}\right)\left(e^{jka}+e^{-j\left(\varphi-kb\right)}\right)}{r^{2}} $$

You should be able to take it from here.