Could the use of a complex rotating phasor for Fourier Series and Transforms be mathematically justified ? Why are we using a complex exponential in the Fourier Transform? Let us suppose there is a vector r in a circle whose tail is at the origin and the head of the vector is at a point p(x,y) and making an angle theta w.r.t to the horizontal x axis. This rotating action can be simply achieved by changing the angle theta and the vector moves around the circle. This rotating action of the vector , will produce the equivalent sine function. Why are we going for a complex exponential in the fourier transform? I understand that the complex exponential also does the same thing and so would any vector r (as mentioned above). Further, the horizontal component of the vector r, along the x axis is rcos(theta) and the vertical component of the vector r is rsin( theta). If you add these , you should get to the head of the vector r. Why would I need to introduce the complex number "i" here , I,e rcos(theta) + ir*sin(theta)? I am unable to understand the use of a complex number here and how is it different from the normal vector "r"(mentioned in the example above), as far as the rotating action is concerned.
I am also aware that re^(i theta) is equal to rcos(theta)+ ir*sin(theta). This is the Euler's complex exponential. It would be helpful if one could explain as to why we use complex number here instead of any other normal vector such as "r " mentioned in the above paragraph. You could as well achieve this rotating action by varying the angle theta of that vector "r" and this rotating action enables you to describe the given f(t) in terms of the sine(theta) or the cosine(theta) respectively
As with many other things, the answer is because we can and because we find it convenient.
Complex exponentials are certainly not the only way to do the job of a Fourier transform. Fourier himself did not use them. Instead, Fourier developed a cosine transform and a sine transform.
One form of a cosine transform is
$$ \int_{-\infty}^\infty \cos(2\pi kt) f(t)\,\mathrm dt $$
and the corresponding sine transform is
$$ \int_{-\infty}^\infty \sin(2\pi kt) f(t)\,\mathrm dt. $$
The cosine and sine functions here are the components of a unit vector that makes and angle $2\pi kt$ counterclockwise from the positive $x$ axis.
Go ahead and use these transforms if you want. Many people do.
The people who use the complex exponential form of Fourier analysis do so because they find it convenient and because it is just as mathematically sound as the cosine and sine transforms.