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The product of (12+34i) and (12−34i) divided by (2−3i) is

Guest Dec 16, 2014

Best Answer 

 #1
avatar+88775 
+5

(12+34i) and (12−34i) divided by (2−3i)

(12 + 34i)(12 - 34i) =

144 - 1156i^2 =

144 - 1156(-1) =

144 + 1156 =

1300

And

1300 / (2-3i)    multiply top and bottom by the conjugate of 2 - 3i

1300(2 + 3i) / [ (2 - 3i)(2 + 3i) =

1300(2 + 3i) / ( 4 - 9i^2)

1300(2 + 3i) / 13 =

2600/13 + 3900i / 13 =

200 + 300i

 

CPhill  Dec 16, 2014
 #1
avatar+88775 
+5
Best Answer

(12+34i) and (12−34i) divided by (2−3i)

(12 + 34i)(12 - 34i) =

144 - 1156i^2 =

144 - 1156(-1) =

144 + 1156 =

1300

And

1300 / (2-3i)    multiply top and bottom by the conjugate of 2 - 3i

1300(2 + 3i) / [ (2 - 3i)(2 + 3i) =

1300(2 + 3i) / ( 4 - 9i^2)

1300(2 + 3i) / 13 =

2600/13 + 3900i / 13 =

200 + 300i

 

CPhill  Dec 16, 2014
 #2
avatar+7188 
0

Good job!!!!

happy7  Dec 16, 2014

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