How do you write the complex number #6-8i# in polar form?
1 Answer
Dec 12, 2016
Explanation:
To express a
#color(blue)"complex number in polar form"#
#"that is " (x,y)to(r,theta)#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))# and
#color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))#
#" where " -pi< theta<=pi#
#"here " x=6" and " y=-8#
#rArrr=sqrt(6^2+(-8)^2)=sqrt(36+64)=10# Now, 6 - 8i is in the 4th quadrant so we must ensure that
#theta"# is in the 4th quadrant.
#theta=tan^-1(-8/6)=-0.927larr" in 4th quadrant"#
#rArr(6,-8)to(10,-0.927)to(10,-53.13^@)#
