Answer
$$A_3 = \sqrt{(A_1)^2+(A_2)^2+2A_1A_2\cos(\delta_1-\delta_2)}$$
$$\delta_3 = \tan^{-1}\big (\frac{A_1\sin(\delta_1) + A_2\sin(\delta_2)}{A_1\cos(\delta_1)+A_2\cos(\delta_2)}\big )$$
Work Step by Step
Given
$$A_3e^{\iota \delta_3} = A_1e^{\iota \delta_1} + A_2e^{\iota \delta_2} $$
Writing, $e^{i\theta} = \cos(\theta)+\iota \sin(\theta)$
$$A_3(\cos(\delta_3)+\iota \sin(\delta_3)) = A_1(\cos(\delta_1)+\iota \sin(\delta_1)) + A_2(\cos(\delta_2)+\iota \sin(\delta_2))$$
$$A_3\cos(\delta_3)+\iota A_3\sin(\delta_3) = (A_1(\cos(\delta_1)+A_2(\cos(\delta_2))+\iota (A_1\sin(\delta_1) + A_2\sin(\delta_2))$$
To Calculate $A_3$, Consider
$$(A_3)^2 = (A_3e^{\iota \delta_3 })\cdot (A_3e^{-\iota \delta_3})$$
$$(A_3)^2 = (A_1e^{\iota \delta_1} + A_2e^{\iota \delta_2})\cdot (A_1e^{-\iota \delta_1} + A_2e^{-\iota \delta_2})$$
$$(A_3)^2 = A_1e^{\iota \delta_1}\cdot A_1e^{-\iota \delta_1}+A_1e^{\iota \delta_1}\cdot A_2e^{-\iota \delta_2}+A_2e^{\iota \delta_2}\cdot A_1e^{-\iota \delta_1}+A_2e^{\iota \delta_2}\cdot A_2e^{-\iota \delta_2}$$
$$(A_3)^2 = (A_1)^2+(A_2)^2+A_1A_2e^{\iota(\delta_1-\delta_2)}+A_1A_2e^{\iota(\delta_2-\delta_1)}$$
$$(A_3)^2 = (A_1)^2+(A_2)^2+A_1A_2e^{\iota(\delta_1-\delta_2)}+A_1A_2e^{-\iota(\delta_1-\delta_2)}$$
$$(A_3)^2 = (A_1)^2+(A_2)^2+A_1A_2(e^{\iota(\delta_1-\delta_2)}+e^{-\iota(\delta_1-\delta_2)})$$
Using the property that, $e^{\iota \theta}+e^{-\iota\theta} = 2\cos(\theta)$
$$(A_3)^2 = (A_1)^2+(A_2)^2+2A_1A_2\cos(\delta_1-\delta_2)$$
$$A_3 = \sqrt{(A_1)^2+(A_2)^2+2A_1A_2\cos(\delta_1-\delta_2)}$$
To calculate $\delta_3$, Consider:
$$\tan(\delta_3) = \frac{A_3 \sin(\delta_3)}{A_3\cos(\delta_3)}$$
$$\tan(\delta_3) = \frac{A_1\sin(\delta_1) + A_2\sin(\delta_2)}{A_1\cos(\delta_1)+A_2\cos(\delta_2)}$$
$$\delta_3 = \tan^{-1}\big (\frac{A_1\sin(\delta_1) + A_2\sin(\delta_2)}{A_1\cos(\delta_1)+A_2\cos(\delta_2)}\big )$$
