Why do Fourier transforms use complex numbers?

You need to ask yourself why we use Fourier transforms. We want to transfer the signal from the space or time domain to another domain - the frequency domain. In this domain, the signal has two "properties" - magnitude and phase. If we want to get only the signal's "power" in a specific frequency bin, we indeed only need to take the absolute value of the Fourier transform, which is real. But, the Fourier transform gives the phase of each frequency as well.

While the first (magnitude's) importance is immediate, the phase is sometimes just as important. For example, for images, most of the information is contained in the phase and NOT in the amplitude. Also, frequency responses (Fourier transforms) are used in digital and analog filters, and the phase plays a major role here as well, especially for audio filters where a linear phase is required: this is what enables an audio filter to process all frequencies and output them without a different delay for each frequency (which will distort the sound - imagine a filter that makes your bass sound come a little before your treble...).

So I hope I convinced you the phase is important as well as the magnitude. And in order to get these two properties, we need something other than just real numbers, we need something with magnitude and phase. Something like a complex number.

You can use the Fourier sine and cosine transforms instead, and these only depend on real functions. This is how Fourier originally worked. \begin{align} f & \sim \frac{1}{\pi}\int_{-\infty}^{\infty}\cos(sx)\int_{-\infty}^{\infty}f(y)\cos(sy)dy \\ & +\frac{1}{\pi}\int_{-\infty}^{\infty}\sin(sx)\int_{-\infty}^{\infty}f(y)\sin(sy)dy. \end{align} The real transforms are analogous to the real Fourier series expansions. This form is cumbersome because of requiring two separate transforms to reconstruct $f$, just as for the real Fourier series. It was realized that the sum of these could be written as $$ \lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}\int_{-R}^{R}f(y)\cos(s(x-y))ds dy \\ =\lim_{R\rightarrow\infty}\frac{1}{\pi}\int_{-\infty}^{\infty}f(y)\frac{\sin(s(x-y))}{x-y}dy \\ =\lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(y)\int_{-R}^{R}e^{is(x-y)}dy \\ = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}e^{isx}\int_{-\infty}^{\infty}e^{-isy}f(y)dyds $$ This was an advance because it simplified the expression to one transform pair instead of two. But you can work with real transforms. The sin transform is $0$ for an even function $f$ and the cos transform is $0$ for an odd function $f$, which corresponds to the traditional Fourier series.