How do you write the complex number in trigonometric form #-1+sqrt3i#?

1 Answer
Aug 25, 2016

#2(cos((2pi)/3)+isin((2pi)/3))#

Explanation:

To convert from #color(blue)"complex to trigonometric form"#

That is #x+yitor(costheta+isintheta)#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|)))" and " color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))#

here x = - 1 and y #=sqrt3#

#rArrr=sqrt((-1)^2+(sqrt3)^2)=sqrt4=2#

Now #-1+sqrt3 i# is in the 2nd quadrant, so we must ensure that #theta# is in the 2nd quadrant.

#theta=tan^-1(-sqrt3)=-pi/3" in 4th quadrant"#

#rArrtheta=(pi-pi/3)=(2pi)/3" in 2nd quadrant"#

#rArr-1+sqrt3i=2(cos((2pi)/3)+isin((2pi)/3))#