Show that Re((b)/(a)) = (bar a b+abar b)/(2|a|^2)…

Show that $$ \text{Re}\left(\frac{b}{a}\right) = \frac{\bar a b+a\bar b}{2|a|^2}; \qquad \text{Im}\left(\frac{b}{a}\right) = \frac{\bar a b-a\bar b}{2i|a|^2}. $$

Step 1 · Rationalize $\dfrac{b}{a}$

$$ \frac{b}{a} = \frac{b}{a}\cdot\frac{\bar a}{\bar a} = \frac{b\bar a}{a\bar a} = \frac{b\bar a}{|a|^2} $$

Step 2 · Real part of a product

For any complex $w$: $\text{Re}(w)=\dfrac{w+\bar w}{2}$. With $w=b\bar a$:

$$ \text{Re}(b\bar a) = \frac{b\bar a+\overline{b\bar a}}{2} = \frac{b\bar a+\bar b a}{2} $$

Step 3 · Assemble $\text{Re}(b/a)$

$$ \text{Re}\left(\frac{b}{a}\right) = \frac{\text{Re}(b\bar a)}{|a|^2} = \frac{b\bar a+a\bar b}{2|a|^2} $$

$$ \boxed{\text{Re}\left(\frac{b}{a}\right) = \frac{\bar a b+a\bar b}{2|a|^2}} $$

Step 4 · Imaginary part of a product

Similarly $\text{Im}(w)=\dfrac{w-\bar w}{2i}$, so:

$$ \text{Im}(b\bar a) = \frac{b\bar a-\overline{b\bar a}}{2i} = \frac{b\bar a-\bar b a}{2i} $$

Step 5 · Assemble $\text{Im}(b/a)$

$$ \text{Im}\left(\frac{b}{a}\right) = \frac{\text{Im}(b\bar a)}{|a|^2} = \frac{b\bar a-a\bar b}{2i|a|^2} $$

$$ \boxed{\text{Im}\left(\frac{b}{a}\right) = \frac{\bar a b-a\bar b}{2i|a|^2}} $$

Cross-check in Cartesian coordinates

Step 6 · Direct computation

Let $a=a_1+ia_2$, $b=b_1+ib_2$. Multiplying by the conjugate of $a$:

$$ \frac{b}{a} = \frac{(b_1+ib_2)(a_1-ia_2)}{a_1^2+a_2^2} = \frac{a_1b_1+a_2b_2}{a_1^2+a_2^2} + i\,\frac{a_1b_2-a_2b_1}{a_1^2+a_2^2} $$

so $\text{Re}(b/a)=\dfrac{a_1b_1+a_2b_2}{|a|^2}$ and $\text{Im}(b/a)=\dfrac{a_1b_2-a_2b_1}{|a|^2}$.

Step 7 · Compute $\bar a b+a\bar b$ and $\bar a b-a\bar b$ directly

$$ \bar a b = (a_1b_1+a_2b_2) + i(a_1b_2-a_2b_1), \qquad a\bar b = (a_1b_1+a_2b_2) + i(a_2b_1-a_1b_2) $$

$$ \bar a b+a\bar b = 2(a_1b_1+a_2b_2), \qquad \bar a b-a\bar b = 2i(a_1b_2-a_2b_1) $$

Step 8 · Confirm both formulas

$$ \frac{\bar a b+a\bar b}{2|a|^2} = \frac{a_1b_1+a_2b_2}{|a|^2} = \text{Re}\left(\frac{b}{a}\right) \quad \checkmark $$

$$ \frac{\bar a b-a\bar b}{2i|a|^2} = \frac{2i(a_1b_2-a_2b_1)}{2i|a|^2} = \frac{a_1b_2-a_2b_1}{|a|^2} = \text{Im}\left(\frac{b}{a}\right) \quad \checkmark $$

Geometric remark · dot and cross products

Viewing $a,b$ as vectors $(a_1,a_2)$, $(b_1,b_2)$ in the plane:

$$ \text{Re}(\bar a b) = a_1b_1+a_2b_2 = \vec a\cdot\vec b \quad\text{(dot product)}, \qquad \text{Im}(\bar a b) = a_1b_2-a_2b_1 = \det(\vec a,\vec b) \quad\text{(2D cross product)} $$

so the two formulas can be written compactly as:

$$ \text{Re}\left(\frac{b}{a}\right) = \frac{\vec a\cdot\vec b}{|a|^2}, \qquad \text{Im}\left(\frac{b}{a}\right) = \frac{\det(\vec a,\vec b)}{|a|^2} $$