+0

# help on question

0
48
7
+1898

can someone help me?

May 1, 2020

#1
+111360
+2

3 -2i   lies  in  Quadrant IV

Multiplying this by i^2  gives us

(3 - 2i) * i^2   =   3i^2 - 2i^3   =  - 3 - 2 (-i)  =   -3 + 2i

This  is a 180°  rotation counter-clockwise

The  new  product  lies in Quadrant II

May 1, 2020
#2
+1898
+1

i jsut went and plotted it and also found that it was in quad 4! bu the rest of the info makes sense as well. thank you!

jjennylove  May 1, 2020
#3
+111360
+1

OK, jenny   !!!

CPhill  May 1, 2020
#4
+1898
0

everything was correct, but it was suppsoed to be 90 degress. I got it wrong. Could you explain ?

jjennylove  May 1, 2020
#5
+111360
+2

Mmmmm...I don't see how they got that  ....but.....I'm not an expert on this !!!!

See the graph  here : https://www.desmos.com/calculator/g6co3qmnlt

We can imagine these points as being in the complex plane

The line  y  =  (-2/3)x   goes through both  points  .....so...they are  separated by 180°

Note  that if we multiply  (3 - 2i)  by i^3   we get   3i^3 - 2i^4  = -3i - 2  =  -2 - 3i   which is a 90° rotation clockwise  and puts the product in Q3

Maybe someone else on here knows why 90° is correct

CPhill  May 1, 2020
#6
+29978
+2

I suspect you dragged 180° into the 2nd box.  However, the second box requires the rotation under multiplication by i, not by i2.  Multiplying by i is a 90° rotation.

Alan  May 1, 2020
#7
+111360
0

Thanks, Alan   !!!!

CPhill  May 1, 2020