# Conversion to Polar form

#### zorro_phu

Joined Jun 14, 2018
20
i don't understand why (j)(j)(-j) when converted to polar form is 1∠90°. How do one get 1∠90°?

#### Papabravo

Joined Feb 24, 2006
17,005
i don't understand why (j)(j)(-j) when converted to polar form is 1∠90°. How do one get 1∠90°?
multiplication is associative so we can write (j)(j)(-j) as ((j)(j))(-j)
(j)(j) = -1 and (-1)(-j) = j
|j| = 1 and ∠j = 90°

Pretty simple and straight forward.

#### MrChips

Joined Oct 2, 2009
24,432
Think of j as a rotation operator that rotates a vector on to the y-axis or j-axis.
Thus (1 + j) is a counter-clockwise rotation 45°.
(0 + j) is a counter-clockwise rotation 90°.
(0 - j) is a clockwise rotation 90°.

Hence (j)(-j) would cancel each other.

(j)(j)(-j) = (j) = (0 + j) = 1∠90°  https://en.wikipedia.org/wiki/Complex_plane

#### WBahn

Joined Mar 31, 2012
26,398
i don't understand why (j)(j)(-j) when converted to polar form is 1∠90°. How do one get 1∠90°?
Remember the definition of j. It is the value that, when squared (multiplied by itself) yields -1.

So (j)(j) = -1 leaving you with (-1)(-j).

Then -x is the same as (-1)(x), so (-j) = (-1)(j), so now you have

(-1)(-j) = (-1)(-1)(j)

Then you know that (-1)(-1) = 1 leaving you with just j.

The only thing left is to realize that j = 1∠90°.

Do you understand why this is the case? If not, we can go into that.

#### MrAl

Joined Jun 17, 2014
8,502
i don't understand why (j)(j)(-j) when converted to polar form is 1∠90°. How do one get 1∠90°?
Hi,

In addition to the other good suggestions here, it would help for you to look up the complex plane and see how that works.