How do you write the complex number #6-8i# in polar form?

1 Answer
Jim G.
Dec 12, 2016

#(10,-0.927)#

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^@)#

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