Reciprocal (Multiplicative Inverse) of a Complex Number

Reciprocal (Multiplicative Inverse) of a Complex Number

For a non-zero complex number $z=a+ib$ (so $a^2+b^2\neq0$), the multiplicative inverse $z^{-1}$ is the complex number satisfying $z\cdot z^{-1}=1$. It is found using exactly the identity from §2.3: multiply and divide by the conjugate $\bar z$, which turns the denominator into the real number $a^2+b^2$:

Writing $|z|^2=a^2+b^2$ (§4), this is the compact formula

Check: $z\cdot z^{-1}=(a+ib)\cdot\dfrac{a-ib}{a^2+b^2}=\dfrac{a^2+b^2}{a^2+b^2}=1$, as required.

The inverse of $z=2+3i$: $|z|^2=4+9=13$, so $z^{-1}=\dfrac{2-3i}{13}=\dfrac2{13}-\dfrac3{13}i$.

Division of complex numbers now follows immediately: dividing by $z_2\neq0$ means multiplying by $z_2^{-1}$, so for $z_1=a+ib$, $z_2=c+id\neq0$,

In practice it is far easier to redo this multiply-by-the-conjugate step each time than to memorise the final formula.

$\dfrac{3+4i}{4-3i}$: multiply top and bottom by $4+3i$ (the conjugate of the denominator): numerator $(3+4i)(4+3i)=12+9i+16i+12i^2=12+25i-12=25i$; denominator $(4-3i)(4+3i)=16+9=25$. So the quotient is $\dfrac{25i}{25}=i$. …