I'm starting to read Ahlfors' Complex Analysis. At the beginning of the book, he uses the fact that $a+ib \neq 0 \implies a^2 +b^2 \neq 0$, but doesn't prove this assertion. This seemed like a pretty obvious statement, so I tried to prove this using only the facts that the book had presented up to this point:
- Two complex numbers are equal if and only if they have the same real part and the same imaginary part.
- Zero is the only number which is at once real and purely imaginary.
Proof:
We see that $$ a+ib \neq 0 \quad \implies \quad a+ib \neq 0 + i0 \quad \implies \quad (a \neq 0) \ \lor \ (b \neq 0) $$ where $\lor$ denotes the inclusive "or". From here we get $2$ cases:
a) If $a \neq 0$, then $a - ib \neq 0$. This is because the only way to have $a - ib = 0$ would be for $a = ib$, but since $ib$ is purely imaginary (by definition) and $a \in \mathbb{R}$, this could only happen if $a=0$ (since $0$ is the only number which is at once real and purely imaginary), contradicting our hypothesis.
b) If $b \neq 0$, then by a similar argument to a) we have that $a - ib \neq 0$.
Since in both cases we get that $a-bi \neq 0$, we see that for any case $$ (a+ib)(a-ib) \neq (0)(0) \quad \implies \quad a^2 + b^2 \neq 0 \qquad \qquad \blacksquare $$
I think my proof is correct, however, I don't like the part where you have to analyze separate cases. I tried to think of a more concise proof (maintaining the restrictions of previous knowledge) but could not think of any other way.
Is my proof correct? And also, does anyone know another way to prove this statement in a more compact manner? Thank you!
It looks to me that you are overcomplicating the issue:
If $a+bi\neq 0$ then either $a\neq 0$ or $b\neq 0$ so clearly $a^2+b^2>0$.