Prove #sqrt(a^2+b^2)e^(iarctan(b/a))=a+bi#?

On a normal coordinate plane, we have coordinate like (1,2) and (3,4) and stuff like that. We can reexpress these coordinates n terms of radii and angles. So if we have the point (a,b) that means we go units to the right, b units up and #sqrt(a^2+b^2)# as the distance between the origin and the point (a,b). I will call #sqrt(a^2 + b^2) = r#

So we have #re^arctan(b/a)#
Now to finish this proof off let's recall a formula.
#e^(itheta) = cos(theta) + isin(theta)#
The function of arc tan gives me an angle which is also theta.
So we have the following equation:
#e^i*arctan(b/a) = cos(arctan(b/a))+sin(arctan(b/a))#

Now lets draw a right triangle.
The arctan of (b/a) tells me that b is the opposite side and a is the adjacent side. So if I want the cos of the arctan(b/a), we use the Pythagorean theorem to find the hypotenuse. The hypotenuse is #sqrt(a^2+b^2)#. So the cos(arctan(b/a)) = adjacent over hypotenuse = #a/sqrt(a^2+b^2)#.

The best part about this is the fact that this same principle applies to sine. So sin(arctan(b/a)) = opposite over hypotenuse = #b/sqrt(a^2+b^2)#.

So now we can re-express our answer as this: #r*((a/sqrt(a^2+b^2))+(bi/sqrt(a^2+b^2)))#.

But remember #r = sqrt(a^2 +b^2)# so now we have: #r *((a/r)+(bi/r))#. The r's cancel, and you are left with the following: #a+bi#

Therefore, #(re^((arctan(b/a))))=a+bi#