complex numbers part 2

O## how do i multiply?

• to find a product (a+bi)(c+di), just expand the brackets the normal way, remembering that $i^2=-1$ (copied word for word from cambridge year 11 textbook 14b summary xd)

how do i divide? part 1 §

• look at $\frac{a+bi}{c+di}$ it is a bit hard to divide normally, so let us learn about the concept of:

conjugate complex numbers! ✨ §

what is a complex conjugate!?

• if $z=a+bi$, then $\overline{z}=a-bi$
• if we multiply $z\text{ and }\overline{z}$, we get $(a+bi)(a-bi)=a^2-abi+abi-b^2{i}^2=a^2+b^2$

how do i divide? part 2 §

• $\frac{a+bi}{c+di}=\frac{a+bi}{c+di}\times \frac{c-di}{c-di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(a+bi)(c-di)}{c^2+d^2}$ and then simplify :D

multiplication inverse §

• to find $z^{-1}$, calculate $\frac{1}{z}$