Question

-6+6 sqrt(3)i Express in trigonometric form.

Answer 1

\[r(\cos \theta + i \sin \theta) = -6 +i6\sqrt{3}\]
sin and cos must be less than 1
r must be 12
\[\rightarrow 12(\frac{1}{2} - i \frac{\sqrt{3}}{2})\]
now solve
\[\cos \theta = \frac{1}{2}\]
\[\sin \theta = -\frac{\sqrt{3}}{2}\]

Answer 2

oh sorry i switched the neg sign
cos = -1/2
sin = sqrt3/2

Answer 3

kk thx

Answer 4

-6+i 6 sqrt(3) = 6 (-1 + i sqrt(3))
6 sqrt((-1)^2 + sqrt(3)^2) exp{i inverse tangent(-sqrt(3)]
12 exp[-i1.0472] or in degrees 6 sqrt(2) exp[-i60]
This commonly called polar form
A cos(x) + iB sin(x)=sqrt(A^2+B^2) exp{iinverse tangent(A/B)

Answer 5

sorry inverse tangent B/A

Answer 6

Sorry
|A + iB| = sqrt(A^2 + B^2)
a angle of inverse tangent B/A
therefore
sqrt(A^2+B^2) (cos (inverse tangent(B/A) + isin(inverse tangent(B/A)
sqrt(A^2+B^2) EXP{i{inverse tangent(B/A)}]