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

## $\sqrt{{a}^{2} + {b}^{2}} {e}^{i \arctan \left(\frac{b}{a}\right)} = a + b i$

Feb 22, 2018

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 $r {e}^{\arctan} \left(\frac{b}{a}\right)$
Now to finish this proof off let's recall a formula.
${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$
The function of arc tan gives me an angle which is also theta.
So we have the following equation:
${e}^{i} \cdot \arctan \left(\frac{b}{a}\right) = \cos \left(\arctan \left(\frac{b}{a}\right)\right) + \sin \left(\arctan \left(\frac{b}{a}\right)\right)$

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 = $\frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$.

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

So now we can re-express our answer as this: $r \cdot \left(\left(\frac{a}{\sqrt{{a}^{2} + {b}^{2}}}\right) + \left(b \frac{i}{\sqrt{{a}^{2} + {b}^{2}}}\right)\right)$.

But remember $r = \sqrt{{a}^{2} + {b}^{2}}$ so now we have: $r \cdot \left(\left(\frac{a}{r}\right) + \left(b \frac{i}{r}\right)\right)$. The r's cancel, and you are left with the following: $a + b i$

Therefore, $\left(r {e}^{\left(\arctan \left(\frac{b}{a}\right)\right)}\right) = a + b i$