Question in proofs review in the complex numbers unit.
I expressed $1/(a+bi) = (a-bi)/(a^2+b^2)$
I then separated the two terms in the denominator to get $a/(a^2+b^2)-bi/(a^2+b^2)$
I then equated the term on the left to $(1/a)$ and the term on the right to $(i/b).$
I rearrange to solve for b in the first equation and am left with $b^2=0$, however b is found in the denominator, thus b can't equal zero.
Is there no solution to this question?
You should just be able to solve the equation and get a condition:
$$\frac{1}{a+ib}=\frac{1}{a}+\frac{i}{b}$$ $$1=1+\frac{ib}{a}+\frac{a}{b}-1=\frac{ib}{a}+\frac{a}{b}$$ $$ab=ib^2+a^2$$ $$0=ib^2+ab+a^2$$ $$b=\frac{-a \pm\sqrt{a^2-4i}}{2i}$$
As pointed out by others, if $a,b\in\mathbb{R}$, then there is no solution.