Question

Complex Numbers

Answer 1

how do you resolve a complex number along a new Re/Im axis ?

Answer 2

@Vocaloid Any ideas? o.o

Answer 3

Was thinking that **if** we want to project \(\mathbf x = <x,y>\) along \(\mathbf u = <u,v>\), the we use the scalar product:

\(\mathbf x \cdot \mathbf u = | \mathbf x | ~ | \mathbf u | \cos \theta\)

so

\(\mathbf x_u = \dfrac{\mathbf x \cdot \mathbf u}{|\mathbf u |} \dfrac{\mathbf u}{|\mathbf u |} = \dfrac{xu + yv}{u^2 + v^2} <u,v> \qquad \star\)

writing as complex numbers:

\(z_1 = x + i y = r_1 e^{i \theta_1}, ~ z_2 = u + i ~ v = r_2 e^{i \theta_2}\)

Resolving \(z_1\) along \(z_2\) ie \((z_1)_2 = |z_1| cos (\theta_1 - \theta_2 ) \dfrac{z_2}{|z_2|}\)

Because \(\dfrac{z_1 }{z_2} = \dfrac{r_1 e^{i \theta_1}}{r_2 e^{i \theta_2}} = \dfrac{r_1 }{r_2 } e^{i \theta_1 - \theta 2}\), then \(\cos (\theta_1 - \theta_2) = \dfrac{r_2}{r_1} \mathbb Re \left( \dfrac{z_1}{z_2} \right) \)

And so:

\((z_1)_2 = |z_1| \dfrac{r_2}{r_1} \mathbb Re \left( \dfrac{z_1}{z_2} \right) \dfrac{z_2}{|z_2|} = \mathbb Re \left( \dfrac{z_1}{z_2} \right) z_2\)


Multiplying that out:

\( = \mathbb Re \left( \dfrac{(x + i y) ( u - i v)}{u^2 + v^2} \right) (u + i v) = \dfrac{x u + y v }{u^2 + v^2} (u + i v) \equiv \star\)


So there are simple scalar and vector product analogues in the complex system