Euler's formula reversed: How does $1.609+1.025i$ become $1.908e^{i32.5^{\circ}}$?

Question

Recently I saw in a youtube video (https://www.youtube.com/watch?v=_h49ilnTmW4&t=508s) where one of the examples was $$e^{i50^{\circ}} + e^{i15^{\circ}} = 1.609+1.025i = 1.908e^{i32.5^{\circ}}$$ My question is: how did he get from $$1.609+1.025i$$ to $$1.908e^{i32.5^{\circ}}$$

Answer 1

When $A,B\in \Bbb R$ and $A, B$ are not both $0$ then the complex number $$Z=A+iB=R(A'+iB')=R(\cos t +i\sin t)=Re^{it}$$ where $R=\sqrt {A^2+B^2},$

and $A'=A/R$ and $B'=B/R,$

and $t$ is any angle such that $\cos t=A'$ and $\sin t=B'.$

Geometrically we can interpret $Z =A+iB$ as the point $(A,B)$ in the plane $\Bbb R^2$. Then $R$ is the Euclidean distance from $Z=(A,B)$ to the origin $O=(0,0),$ and $t$ is the angle between the line-segment $ZO$ and the half-axis $[0,\infty)\times \{0\},$ with $t>0$ if $B>0$ and $t<0$ if $B<0.$... And with $t=0$ if $A>0=B,$ while $t=-\pi$ if $A<0=B.$

Of course we can change $t$ to $t+2n\pi$ for any integer $n$ without affecting $Z.$

It is usually preferable to work in radians in pure maths instead of degrees. A radian is not even a "unit" since $\pi$ radians is the same thing as the number $\pi.$

Answer 2

since the two complex numbers have the same magnitude, that being $1,$ we get a perfect rhombus with the four numbers $0, e^{i 15^\circ}, e^{i 50^\circ}$ and the sum you are asked about. The sum is along the angle bisector and must be a multiple of $e^{i 32.5^\circ}$ since $(50+15)/2= 65/2 = 32.5$

Neither the short diagonal nor the long diagonal of the rhombus is particularly nice as far as magnitude. However, we can find the long diagonal with the Law of Cosines, $$ c^2 = 1 + 1 - 2 \cos 145^\circ \approx 1 + 1 + 2 \cdot 0.819152044 \approx 3.638304089, $$ so the magnitude is $$ c \approx 1.907433902 $$