Real and Imaginary Part Identities

Let $z = a + bi$, so $\bar z = a - bi$. Add and subtract:

• $z + \bar z = (a + bi) + (a - bi) = 2a = 2\text{Re}(z)$, so $\text{Re}(z) = \tfrac{z + \bar z}{2}$.
• $z - \bar z = (a + bi) - (a - bi) = 2bi = 2i\,\text{Im}(z)$, so $\text{Im}(z) = \tfrac{z - \bar z}{2i}$.

Worked through the hook: $z = 3 + 4i$, $\bar z = 3 - 4i$. $z + \bar z = 6 = 2 \cdot 3$, so divide by $2$ to get $\text{Re}(z) = 3$. $z - \bar z = 8i = 2i \cdot 4$, so divide by $2i$ to get $\text{Im}(z) = 4$. The denominator for Re is $2$; for Im it is $2i$ — that is the trap.

The other half of the toolkit:

• Modulus identity: $z\bar z = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2$. Always a non-negative real number.
• Multiplicative modulus: $|z_1 z_2|^2 = (z_1 z_2)\overline{(z_1 z_2)} = z_1 z_2 \bar z_1 \bar z_2 = (z_1 \bar z_1)(z_2 \bar z_2) = |z_1|^2 |z_2|^2$. Taking positive square roots: $|z_1 z_2| = |z_1||z_2|$.

$\text{Re}(z) = (z + \bar z)/2$ — divisor is $2$ · $\text{Im}(z) = (z - \bar z)/(2i)$ — divisor is $2i$, not $2$ · $z\bar z = |z|^2$ — replace $|z|^2$ by $z\bar z$ in proofs · $|z_1 z_2| = |z_1||z_2|$ — proved via $|z|^2 = z\bar z$ and $\overline{z_1 z_2} = \bar z_1 \bar z_2$

Pause — copy $\operatorname{Re}(z) = (z+\bar{z})/2$, $\operatorname{Im}(z) = (z-\bar{z})/(2i)$, and the product-modulus proof $|z_1 z_2|^2 = z_1 z_2 \overline{z_1 z_2} = |z_1|^2|z_2|^2$ into your book.

We just saw that $\operatorname{Re}(z) = (z+\bar{z})/2$ and $\operatorname{Im}(z) = (z-\bar{z})/(2i)$, and that $|z_1 z_2| = |z_1||z_2|$ is proved via $|z|^2 = z\bar{z}$. That raises a question: how do you use these identities inside a proof — for example, to show that some expression is purely real or purely imaginary? This card answers it → $w$ real $\Leftrightarrow w = \bar{w}$; $w$ purely imaginary $\Leftrightarrow w + \bar{w} = 0$.

The identities $\text{Re}(z) = \tfrac{z+\bar z}{2}$, $\text{Im}(z) = \tfrac{z-\bar z}{2i}$, $z\bar z = |z|^2$ and $|z_1 z_2| = |z_1||z_2|$ work together. Three common patterns:

• "Show $w$ is real." Show $w = \bar w$. (Because $w - \bar w = 0$ means $\text{Im}(w) = 0$.)
• "Show $w$ is purely imaginary." Show $w + \bar w = 0$. (Because then $\text{Re}(w) = 0$.)
• "Prove $|z_1 z_2| = |z_1||z_2|$." Square both sides: $|z_1 z_2|^2 = (z_1 z_2)\overline{(z_1 z_2)} = z_1 \bar z_1 \cdot z_2 \bar z_2 = |z_1|^2 |z_2|^2$.

$w$ real $\Leftrightarrow w = \bar w$ — strategy for "show $w$ is real" · $w$ purely imaginary $\Leftrightarrow w + \bar w = 0$ — strategy for "show purely imaginary" · To prove modulus identities, square them and use $|z|^2 = z\bar z$ · $\text{Im}(z)$ is the real number $b$, not $bi$

Pause — copy the two reality tests ($w$ real $\Leftrightarrow w=\bar{w}$; $w$ pure imaginary $\Leftrightarrow w+\bar{w}=0$), the strategy for modulus proofs (square and use $|z|^2 = z\bar{z}$), and the note that $\operatorname{Im}(z) = b$ not $bi$ into your book.