Use square roots and conjugates

Understand conjugation as a reflection

Recall the surd pair $a+\sqrt b$ and $a-\sqrt b$.

The notes motivate the complex analogue by writing $2+3i=2+3\sqrt{-1}$, whose conjugate is $2-3\sqrt{-1}=2-3i$. For real $x,y$, define

$$z=x+iy\quad\Longrightarrow\quad z^*=x-iy.$$

The real part stays the same; the imaginary coefficient changes sign.

The source pair $1+3i$ and $1-3i$ lies at $(1,3)$ and $(1,-3)$: reflecting the point and its position vector in the real axis gives the conjugate.

A real number is unchanged because its imaginary coefficient is zero.

For the calculator illustrated in the source, use MODE → a+bi, then MATH → CMPLX → 1:conj(. The screen gives $\operatorname{conj}(1+3i)=1-3i$ and $\operatorname{conj}(-1-4i)=-1+4i$.

Derive the first eight conjugate properties

Let $z=x+iy$ with real $x,y$. These are the source rows (a)–(h), including their derivations.

(a) Definition. $z^*=x-iy$ by definition.

(b) Conjugate twice.

$$(z^*)^*=(x-iy)^*=x+iy=z.$$

(c) Product with the conjugate.

$$\begin{aligned}zz^*&=(x+iy)(x-iy)\\&=x^2-ixy+ixy-(iy)^2\\&=x^2-i^2y^2\\&=x^2+y^2.\end{aligned}$$

(d) A real scalar passes through conjugation. For $k\in\mathbb R$,

$$\begin{aligned}(kz)^*&=[k(x+iy)]^*\\&=(kx+iky)^*\\&=kx-iky\\&=k(x-iy)\\&=kz^*.\end{aligned}$$

The assumption that $k$ is real matters; it is not a rule for an arbitrary complex scalar without conjugating that scalar too.

(e) Add conjugates to recover the real part.

$$z+z^*=(x+iy)+(x-iy)=2x=2\operatorname{Re}(z).$$

(f) Subtract conjugates to recover the imaginary coefficient.

$$z-z^*=(x+iy)-(x-iy)=2iy=i[2\operatorname{Im}(z)].$$

The factor $i$ is essential: $\operatorname{Im}(z)$ itself is a real number.

(g) Conjugation preserves addition. With real components $x_1,y_1,x_2,y_2$,

$$\begin{aligned}(z_1+z_2)^*&=[(x_1+iy_1)+(x_2+iy_2)]^*\\&=[(x_1+x_2)+i(y_1+y_2)]^*\\&=(x_1+x_2)-i(y_1+y_2)\\&=(x_1-iy_1)+(x_2-iy_2)\\&=z_1^*+z_2^*.\end{aligned}$$

(h) Extend the addition rule to a finite sum. Apply (g) repeatedly, separating the last term at each stage:

$$\begin{aligned}(z_1+\cdots+z_n)^*&=(z_1+\cdots+z_{n-1})^*+z_n^*\\&=(z_1+\cdots+z_{n-2})^*+z_{n-1}^*+z_n^*\\&\;\vdots\\&=z_1^*+z_2^*+\cdots+z_n^*.\end{aligned}$$

Complete the conjugate product rules

For $z_1=x_1+iy_1$, $z_2=x_2+iy_2$ with real components, the remaining source properties are (i)–(k).

(i) Conjugation preserves multiplication. Expand, conjugate the collected form, and regroup into the two conjugates:

$$\begin{aligned}(z_1z_2)^*&=[(x_1+iy_1)(x_2+iy_2)]^*\\&=[x_1x_2+ix_1y_2+ix_2y_1+i^2y_1y_2]^*\\&=[(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1)]^*\\&=(x_1x_2-y_1y_2)-i(x_1y_2+x_2y_1)\\&=x_1x_2-ix_1y_2-y_1y_2-ix_2y_1\\&=x_1(x_2-iy_2)+(-i)^2y_1y_2-ix_2y_1\\&=x_1(x_2-iy_2)+(-i)y_1(-iy_2+x_2)\\&=(x_1-iy_1)(x_2-iy_2)\\&=z_1^*z_2^*. \end{aligned}$$

Here $(-i)^2=-1$ is the key regrouping identity.

(j) Extend to a finite product. Apply (i) repeatedly:

$$\begin{aligned}(z_1z_2\cdots z_n)^*&=(z_1\cdots z_{n-1})^*z_n^*\\&=(z_1\cdots z_{n-2})^*z_{n-1}^*z_n^*\\&\;\vdots\\&=z_1^*z_2^*\cdots z_n^*. \end{aligned}$$

(k) Conjugate a positive integer power. Regard $z^n$ as a product of $n$ identical factors:

$$\begin{aligned}(z^n)^*&=(\underbrace{z\cdot z\cdots z}_{n\text{ factors}})^*\\&=\underbrace{z^*\cdot z^*\cdots z^*}_{n\text{ factors}}\\&=(z^*)^n. \end{aligned}$$