
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