Re{...} function specification

76 Views Asked by At

I am doing a proof that uses the $Re\{\}$ function (the function that outputs the Real part of a complex number $a + bi$). Along the way I realized that I am not sure how the function behaves when it comes to multiplied inputs. Specifically, is

$Re\{\phi_1 • \phi_2\}$ where $\phi_1, \phi_2$ are complex numbers equivalent to $Re\{\phi_1\} • Re\{\phi_2\}?$

2

There are 2 best solutions below

0
On BEST ANSWER

Since $$(a+bi)(c+di) = (ac-bd)+(ad+bc)i$$ we have that $$\operatorname{Re}(zw) = \operatorname{Re}(z)\operatorname{Re}(w) - \operatorname{Im}(z)\operatorname{Im}(w).$$

0
On

No, it's not. This is how complex multiplication works: $$ (a+b\mathrm i)\cdot(c+d\mathrm i) = (ac-bd)+(ad+bc)\mathrm i. $$ Hence, when $z,w$ are complex numbers, we have $$ \operatorname{Re}(z\cdot w) = \operatorname{Re}(z)\cdot\operatorname{Re}(w)-\operatorname{Im}(z)\cdot\operatorname{Im}(w). $$