What is the polar form of #( -4,32 )#?

The rectangular point #(-4,32)# is in the form #(x,y)#.

Polar points are in the form #(r,theta)#. See the attached image for what this means:

To find #r#, we effectively need to find the hypotenuse of the right triangle with legs of #x# and #y#.

Thus, #r=sqrt(x^2+y^2)#. Here, this becomes

#r=sqrt((-4)^2+(32)^2)=sqrt(4^2+32^2)=sqrt(4^2+4^2(8^2))=sqrt(4^2(1+8^2))=4sqrt65#

Even though the point #(-4,32)# is in Quadrant #"II"#, the value of #r# is a magnitude and is still positive.

To find #theta#, we first need to write some statement involving #theta# given the information that we know.

Looking at the image, we have #theta#, the side opposite #theta#, and the side adjacent to #theta# in a right triangle. Thus, we can say:

#tantheta="opposite"/"adjacent"=y/x#

Solving for #theta#:

#theta=tan^-1(y/x)#

Using our known values:

#theta=tan^-1(32/(-4))=tan^-1(-8)=-1.44644133#

Note, however, that this is a negative value and that #-1.44644133gt-pi/2#, so this is really an angle in Quadrant #"IV"#.

To find the value of this angle in Quadrant #"II"#, we know that it will be #pi# minus the magnitude of the angle we determined.

That is,

#theta=pi-1.44644133=1.69515132#

So, our point is:

#(r,theta)=(4sqrt65,1.69515132)#