Transform $r_1$ into trigonometric form

Question

Let $a, b ∈ \mathbb{R}$ be fixed numbers. Assume the characteristic equation homogeneous recursive identity (4.1) $$s_n=as_{n-1} + bs_{n-2}$$ where $$n \geq 2 $$ there are no real roots (i.e. the discriminant ∆ of this equation is negative).
(a) Show that the characteristic equation $r^2-ar-b=0$ has two roots $r_1, r_2$ , which are conjugate complex numbers.
(b) From (a) it follows that the characteristic roots $r_1, r_2$ of the identity (4.1) can be written in trigonometric form: $$r_1 = ρ (cos α + isin α), r_2 = ρ (cos α - isin α)$$ where ρ> 0 and α ∈ [0, 2π). Prove that the sequences $(u_n)_{n∈N}$ and $(v_n)_{n∈N}$ with the terms (real) defined by formulas $$u_n=ρ^ncos(nα), v_n=ρ^nsin(nα)$$for $n∈N_0$ satisfy the identity (4.1) and are linearly independent, so they form the basis of a linear space $e_{a,b}$ composed of two strings with real terms. Tip: The string satisfying the identity (4.1) is the sequence ($r^n_1)_{n∈N_0}$. Use de Moivre's formula to write $(r^n_1)$ in the form trigonometric.
a) $$r^2-ar-b$$ $$∆=a^2+4b<0$$ $$a^2<-4b$$ $$\sqrt\Delta=\sqrt {a^2+4b}=\sqrt {i^2|a^2+4b|}=i\sqrt {|a^2+4b|}$$ $$r_1=\frac{a-\sqrt{a^2+4b}}{2}=\frac{a-i\sqrt{|a^2+4b|}}{2}$$ $$r_1=\frac{1}{2}a-i \frac{1}{2} \sqrt{|a^2+4b|}$$ $$r_2=\frac{1}{2}a+i\frac{1}{2} \sqrt{|a^2+4b|}$$ $$Re(r_1)=\frac{1}{2}a=Re(r_2)$$ $$Im(r_1)=- \frac{1}{2} \sqrt{|a^2+4b|}$$ $$Im(r_2)=\frac{1}{2} \sqrt{|a^2+4b|}$$ $$Im(r_1)=-Im(r_2)$$ $$r_1={\over r_2}$$ $${\over r_1}=r_2$$ $$Sn=A(r_1)^n+B(r_2)^n$$ I have a problem with b) with transforming $r_1$ into trigonometric form, could anyone help? I am thinking about that for two days and do not know what to do now.