How to find the argument of the dot product of two complex vectors?

Question

I have two complex vectors $\mathbb{a}$ and $\mathbb{b}$ I want to find $\arg(c)$ where $$c=\langle\mathbb{a},\mathbb{b}\rangle.$$ That is $c$ is the inner product of the two vectors. Thank you.

Answer 1

Hint:

since the argument of a complex number $z$ is $$ \theta=\arctan \frac{Im (z)}{Re (z)} $$ you can use the fact that, with the definition of the inner product,

$$ \langle\mathbb{a},\mathbb{b}\rangle=\sum_{i=1}^n \overline{a_i}b_i $$

so that $$ \overline{\langle\mathbb{a},\mathbb{b}\rangle}=\sum_{i=1}^n \overline{b_i}a_i=\langle\mathbb{b},\mathbb{a}\rangle $$ we have:

$$Im (\langle\mathbb{a},\mathbb{b}\rangle)=\frac{1}{2i}(\langle\mathbb{a},\mathbb{b}\rangle-\langle\mathbb{b},\mathbb{a}\rangle)$$

and

$$Re (\langle\mathbb{a},\mathbb{b}\rangle)=\frac{1}{2}(\langle\mathbb{a},\mathbb{b}\rangle+\langle\mathbb{b},\mathbb{a}\rangle)$$

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putting in the definition of $\theta$ we can find:

$$ \theta=\arctan \left(-i\frac{\langle\mathbb{a},\mathbb{b}\rangle)-\langle\mathbb{b},\mathbb{a}\rangle)}{\langle\mathbb{a},\mathbb{b}\rangle)+\langle\mathbb{b},\mathbb{a}\rangle)} \right) = \arctan \left(-i\frac{\langle\mathbb{a},\mathbb{b}\rangle)^2-\langle\mathbb{b},\mathbb{a}\rangle)^2}{|\langle\mathbb{a},\mathbb{b}\rangle)+\langle\mathbb{b},\mathbb{a}\rangle)|^2} \right) $$