Question

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

`sqrt(a^2+b^2)e^(iarctan(b/a))=a+bi`

Answer 1

In Explanation

Explanation:

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`