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

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

##### 1 Answer

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

So we have

Now to finish this proof off let's recall a formula.

The function of arc tan gives me an angle which is also theta.

So we have the following equation:

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

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

So now we can re-express our answer as this:

But remember

Therefore,