Question 503933: 3-9i/6-9i in complex form
Answer by nic0le116(25)   (Show Source):
You can put this solution on YOUR website! We need to get rid of the i's in the denominator.
You do this by multiplying the top and bottom by the conjugate of the denominator.
The conjugate of 6-9i is 6+9i.
You just need to change the sign of the second number.
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So lets multiply the numerator first.
(3-9i)(6+9i)= 18+27i-54i+81i^2
REMEMBER- i^2 is equal to -1. So replace the i^2 with -1
Multiply 81 by -1
= 18-27i-81
-63-27i <----------THIS IS OUR NUMERATOR FOR NOW
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Now, lets multiply the conjugate by the denominator
Remember the goal is to get rid of all i's in the denominator.
(6-9i)(6+9i) =36+54i-54i-81i^2
(i's cancel and we're left with only an i^2 which we now know needs to be replaced with negative 1)
=36-(81*-1)
=36+81
=117 <-----------------THIS IS OUR DENOMINATOR FOR NOW.
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Now let's put them together in complex form which is a+bi.
This simply means numbers on one side and all numbers attatched to an i on the other
NOT IN COMPLEX FORM ANSWER= -63-27i/117
Since out denominator is 117, all numbers need to be over 117, but broken up so i's are to the right.
ANSWER IN COMPLEX FORM= (-63/117)-(27i/117)
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