What is the trigonometric form of (1-3i) (1−3i)?
1 Answer
May 2, 2016
Explanation:
Given a complex number z = x + iy , then in trig.form it is written
z =
r(costheta + isintheta)r(cosθ+isinθ) where
|z|=|x+iy|=r=sqrt(x^2+y^2)|z|=|x+iy|=r=√x2+y2 and
theta=tan^-1(y/x)θ=tan−1(yx) here x = 1 and y = - 3
rArr r=sqrt(1^2+(-3)^2)=sqrt10⇒r=√12+(−3)2=√10 and
theta=tan^-1(-3)=-1.25" radians "θ=tan−1(−3)=−1.25 radians
rArr(1-3i)=sqrt10(cos(-1.25)+isin(-1.25))⇒(1−3i)=√10(cos(−1.25)+isin(−1.25))
