SOLUTION: Does anyone know the steps for this type of question .I need help .thanks. 1-i^34 divided by 3 square root -1 +5 (note that only -1 is under the square root and 3 is in front of

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Question 725615: Does anyone know the steps for this type of question .I need help .thanks.
1-i^34 divided by 3 square root -1 +5 (note that only -1 is under the square root and 3 is in front of it )

Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
1-i^34 divided by 3 square root -1 +5
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(1-i^(34))/(3i + 5)
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Simplify the numerator:
i^34 = (i^4)^8*i^2 = 1^8*(-1) = -1
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Your Problem:
= (1-1)/(3i + 5)
= 0/(5+3i)
--
= 0
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Cheers,
Stan H.
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Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
1-i^34 divided by 3 square root -1 +5 (note that only -1 is under the square root and 3 is in front of it )
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i^34 = i^32*i^2 = 1*-1 = -1
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(1-i^34)/(3i+5)
= 2/(5 + 3i)
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Get the i out of the denominator
Multiply NUM and DEN by the conjugate of the DEN 5 - 3i
= 2(5-3i)/(5^2 + 9i^2)
= 2(5-3i)/16
= (5-3i)/8
or
5/8 - i*(3/8)