Rectangular to Polar Form

[nycmath]

See attachment. Enjoy.

[Eigenvalue]

(3,√3)
r=[tex]\sqrt{3²+(√3)²}[/tex]=[tex]\sqrt{12}[/tex]=2√3
tan(theta)=[tex]\frac{3}{√3}[/tex]=√3
theta=60°
3(cos 60°+sin60°)
3(cos(π/3)+sin(π/3))

[Eigenvalue]

(-1,-1)
r=[tex]\sqrt{(-1)²+(-1)²}[/tex]=[tex]\sqrt{2}[/tex]
tan(theta)=[tex]\frac{-1}{-1}[/tex]=1
theta=π/2or 3π/2
theta=3π/2
[tex]\sqrt{2}[/tex](cos(3π/2)+sin(3π/2))

[nycmath]

(3,√3)
r=[tex]\sqrt{3²+(√3)²}[/tex]=[tex]\sqrt{12}[/tex]=2√3
tan(theta)=[tex]\frac{3}{√3}[/tex]=√3
theta=60°
3(cos 60°+sin60°)
3(cos(π/3)+sin(π/3))

What is the formula to do this with?

[nycmath]

(-1,-1)
r=[tex]\sqrt{(-1)²+(-1)²}[/tex]=[tex]\sqrt{2}[/tex]
tan(theta)=[tex]\frac{-1}{-1}[/tex]=1
theta=π/2or 3π/2
theta=3π/2
[tex]\sqrt{2}[/tex](cos(3π/2)+sin(3π/2))

The 7 train is stuck at 33rd Street. So, why not check out math10? Nice work here.