How do you convert #2 - 2i# to polar form?

1 Answer
Jim G.
Aug 4, 2016

#(2,-2)to(2sqrt2,-pi/4)#

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 = 2 and y = -2

#rArrr=sqrt(2^2+(-2)^2)=sqrt8=2sqrt2#

Now 2 - 2i is in the 4th quadrant, hence we must ensure that #theta# is in the 4th quadrant.

#theta=tan^-1((-2)/2)=tan^-1(-1)=-pi/4" in 4th quadrant"#

#rArr2-2i=(2,-2)to(2sqrt2,-pi/4)#

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