How do you write #(2)-(2isqrt3)# in trigonometric form?

1 Answer
Aug 26, 2016

#4(cos(pi/3)-isin(pi/3))#

Explanation:

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

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

#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 = 2 and #y=-2sqrt3#

#rArrr=sqrt(2^2+(2sqrt3)^2)=sqrt(4+12)=sqrt16=4#

Now #2-2sqrt3 i# is in the 4th quadrant so we must ensure that #theta# is in the 4th quadrant.

#theta=tan^-1((-2sqrt3)/2)=tan^-1(-sqrt3)#

#=-pi/3" in 4th quadrant"#

#rArr2-2sqrt3 i=4(cos(-pi/3)+isin(-pi/3))#

which can also be expressed as #4(cos(pi/3)-isin(pi/3))#

Since #color(red)(|bar(ul(color(white)(a/a)color(black)(cos(-theta)=costheta" and " sin(-theta)=-sintheta)color(white)(a/a)|)))#