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0
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avatar+1982 

 

can someone help me?

 May 1, 2020
 #1
avatar+112185 
+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

 

cool cool cool

 May 1, 2020
 #2
avatar+1982 
+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
avatar+112185 
+1

OK, jenny   !!!

 

cool cool cool

CPhill  May 1, 2020
 #4
avatar+1982 
0

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

jjennylove  May 1, 2020
 #5
avatar+112185 
+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

 

cool cool cool

CPhill  May 1, 2020
 #6
avatar+31296 
+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
avatar+112185 
0

Thanks, Alan   !!!!

 

 

cool cool cool

CPhill  May 1, 2020

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