SOLUTION: How do you solve 4+9i/12i into an expression of standard form?

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Question 250429: How do you solve 4+9i/12i into an expression of standard form?
Found 2 solutions by stanbon, jsmallt9:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How do you solve (4+9i)/(0+12i) into an expression of standard form?
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Multiply numerator and denominator by (0-12i) to get:
[(4+9i)(0-12i)]/[(0+12i)(0-12i)]
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[0_-48i+0i+108]/[144]
= [108-48i]/144
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Cheers,
Stan H.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%284%2B9i%29%2F12i
Since a + bi has two terms, we will start by splitting this fraction into two terms:
4%2F12i%2B9i%2F12i
(Think of the above as "un-adding".) Next we can simplify the fractions:
1%2F3i%2B3%2F4
Since i = sqrt%28-1%29 we do not want it in the denominator. We can rationalize the first fraction by multiplying the top and bottom by i:
%281%2F3i%29%28i%2Fi%29%2B3%2F4
i%2F3i%5E2+%2B+3%2F4
Since i%5E2+=+-1:
i%2F3%28-1%29+%2B+3%2F4
i%2F%28-3%29+%2B+3%2F4
We want the real term, a, in front:
3%2F4+%2B+i%2F%28-3%29
We want b*i not i divided by something so, since dividing by -3 is the same as multiplying by -1/3:
3%2F4+%2B+%28%28-1%29%2F3%29i