Parviz F. answered  02/18/14
Tutor
                    
        4.8
        (4)
            Mathematics professor at Community Colleges
a)
1/9 = 1/9 e0(i)  = 1/9 e0
b )
 6 + 6i
  r = √( 6^2 + 6^2) = 6√2
θ = Tan -1( 1) = Π/ 4
6 + 6i = 6 √2 e iΠ/4
  c )
     c)
       -3 - 3i
       r = √ (( -3) ^2  + ( -3) ^2 ) = 3√2
            θ = tan-1( -3/ -3 ) = 5Π/4
       - 3 - 3 i = 3√2 e i5Π/4