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

##### 1 Answer

#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)|)))#