How do you convert #4+4i # to polar form?

1 Answer
Jim G.
Jul 13, 2016

#(4sqrt2,pi/4)#

Explanation:

To convert from #color(blue)"cartesian to polar form"#

That is (x ,y)#to(r,theta)# use

#color(orange)"Reminder"#

#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)|)))#

Now 4 + 4i is in the 1st quadrant ,hence we must ensure that #theta# is an angle in the 1st quadrant.

here x = 4 and y = 4

#rArrr=sqrt(4^2+4^2)=sqrt32=4sqrt2#

and #theta=tan^-1(4/4)=tan^-1(1)=pi/4" angle in 1st quadrant"#

#rArr(4,4)=(4sqrt2,pi/4)" in polar form"#

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