Question about proving multiplicative inverse of complex numbers

Question

If I want to prove for complex numbers $z_1 \neq 0$ there exists a multiplicative inverse $z_2$ such that $z_1z_2=1$.I am having trouble deciding between using two different proofs.

Proof#1:

Let $z_1=a+bi,z_2=c+di$

Then to find the multiplicative inverse of $z_1$

Let $(a+bi)(c+di)=1 \implies (ac-bd)+(bc+ad)i=1$

This yields the system of equations

$ac-bd=1$

$bc+ad=0 \implies bc=-ad \implies c=\frac{-ad}{b}$

Now at this point I am questioning this method since $a$ could be a real number(with imaginary part $b=0$).

Another method:

I feel like this method is more appropriate

$(a+bi)(c+di)=1$, Since $a+bi \neq 0$,

$c+di=\frac{1}{a+bi}$

Then I would finish the proof by multiplying the numerator and denominator by the conjugate to get $c=\frac{a}{a^2+b^2}$ and $d=\frac{-b}{a^2+b^2}$

Then I would show $(a+bi)(\frac{a-bi}{a^2+b^2})=1$

My question is why would anyone use the first method? Is there some explanation for why we can divide by $b$? To me it is not evident whether $a$ or $b$ or both are nonzero, so how can we get away with using the first method?

Answer 1

$a$ and $b$ cannot both be $0$. Multiply $a+bi$ by $a-bi$. You will see that the result is a real non-zero number. Call this number $r$. Then the multiplicative inverse of $a + bi$ is
$$\frac ar - \frac br i.$$

Answer 2

If $b\ne0$ you can substitute $c=-ad/b$ into $1=ac-bd=-d(a^2+b^2)/b$ so$$d=-b/(a^2+b^2),\,c=a/(a^2+b^2)$$as required. If $b=0$ we can't get that formula for $c$ in the first place, but the problem simplifies to inverting the real $a$ viz.$$ac=1,\,ad=0\implies c=1/a,\,d=0.$$Alternatively, we can solve the simultaneous equations $ac-bd=1,\,bc+ad=0$ for $c,\,d$ by matrix algebra.