How do you write (2)-(2isqrt3)(2)−(2i√3) in trigonometric form?
1 Answer
4(cos(pi/3)-isin(pi/3))4(cos(π3)−isin(π3))
Explanation:
To convert from
color(blue)"complex to trigonometric form"complex to trigonometric form That is
x+yitor(costheta+isintheta)x+yi→r(cosθ+isinθ)
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 thattheta 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)|)))
