Multiplication of Complex Numbers

Multiplication of Complex Numbers

For $z_1=a+ib$ and $z_2=c+id$, multiplication is carried out exactly like multiplying two binomials, and then $i^2$ is replaced by $-1$:

so, collecting real and imaginary parts,

Because $ac-bd$ and $ad+bc$ are real numbers whenever $a,b,c,d$ are, the product is again a complex number: $\mathbb{C}$ is closed under multiplication.

$(3+2i)(1-4i)=3(1)+3(-4i)+2i(1)+2i(-4i)=3-12i+2i-8i^2=3-10i+8=11-10i$, using $-8i^2=-8(-1)=8$.

Properties of multiplication:

1. Commutative: $z_1z_2=z_2z_1$ — expanding both sides gives $(ac-bd)+i(ad+bc)$ either way.
2. Associative: $(z_1z_2)z_3=z_1(z_2z_3)$, verified by expanding both sides using the definition above.
3. Distributive over addition: $z_1(z_2+z_3)=z_1z_2+z_1z_3$.
4. Multiplicative identity: $z\cdot1=z$ for every $z$, taking $1=1+0i$.

Powers of $i$. Since $i^2=-1$, repeated multiplication by $i$ cycles through only four values: $i^1=i$, $i^2=-1$, $i^3=i^2\cdot i=-i$, and $i^4=i^2\cdot i^2=(-1)(-1)=1$, after which the pattern $i,-1,-i,1$ repeats forever. For any integer $n$, write $n=4q+r$ with remainder $0\le r<4$ (ordinary division of $n$ by $4$); then $i^n=(i^4)^q\cdot i^r=1^q\cdot i^r=i^r$, so evaluating any power of $i$ reduces to reading off one of just four values. …