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2006-10-02 10:49:53 · 3 answers · asked by jose u 1 in Science & Mathematics Mathematics

3 answers

(2-3i)/(-3+i)
(2-3i)(-3-i)/(9+1)
(-6-2i+9i-3)/10
(-9+7i)/10

10/(-9+7i)
10(-9-7i)/(81+49)
(-90-70i)/130

2006-10-02 10:55:41 · answer #1 · answered by Greg G 5 · 0 0

(2-3i) / (-3+i)

usually you want to get rid of the imaginary element in the denominator

in this case you do that by multiplying by to and bottom of fraction by (-3-i)

(2-3i)(-3-i)/(-3+i)(-3-i)

-6+7i-3 / 9+1

-9+7i / 10


even if I messed that up somewhere, perhaps you get the idea

2006-10-02 17:55:46 · answer #2 · answered by enginerd 6 · 0 0

Buckfush

2006-10-02 17:51:27 · answer #3 · answered by joxer 2 · 0 0

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