How do you convert #3+2i# to polar form?

1 Answer
Jim G.
Sep 8, 2016

#(sqrt13,0.588)#

Explanation:

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

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

#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)|)))#

here x = 3 and y = 2

#rArrr=sqrt(3^2+2^2)=sqrt13#

Now, 3 + 2i, is in the 1st quadrant so we must ensure that #theta# is in the 1st quadrant.

#theta=tan^-1(2/3)=0.588" radians"larr" in 1st quadrant"#

Thus #(3,2)to(sqrt13,0.588)to(sqrt13,33.69^@)#

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