What is the polar form of #( -4,5 )#?

1 Answer
Dec 17, 2015

The polar form of (-4,5) has #sqrt(41)# as module and #arccos(-4/sqrt(41))# as argument.

Explanation:

You can use Pythagoras theorem or the complex numbers. I'm gonna use the complex numbers because it is simpler to write down and to explain as I always do that and english is not my mother language.

By identifying #RR^2# as the complex plan #CC#, #(-4,5)# is the complex number #-4 + 5i#. Its module is #abs(-4+5i) = sqrt(5^2 + (-4)^2) = sqrt(41)#.

We now need the argument of this complex number. We know its module, so we can write that #-4+5i = sqrt41(-4/sqrt41 + i5/sqrt41)#.

We know that when we factorize by the module, we get the cosine and the sine of a real number. It means that #EE alpha in RR# such that #cos(alpha) = -4/sqrt41# and #sin(alpha) = 5/sqrt(41)#. So #alpha = arccos(-4/sqrt(41))# which is the argument of (-4,5).