SOLUTION: Find modulus of complex number z= (((3-4j)(3j-2))/-j)
i have tried
=((9j-6-12j^2+8j)/-j)
=((17j+6)/-j)
modulus = (sqrt(17^2 +6^2))/-j = (sqrt(325))/-j
Is this correct
Question 1136493: Find modulus of complex number z= (((3-4j)(3j-2))/-j)
i have tried
=((9j-6-12j^2+8j)/-j)
=((17j+6)/-j)
modulus = (sqrt(17^2 +6^2))/-j = (sqrt(325))/-j
Is this correct Found 2 solutions by rothauserc, greenestamps:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! z = (((3-4j)(3j-2))/-j) =
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perform complex number division
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Note this problem uses j instead of i, therefore j = square root(-1) and j^2 = -1
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multiply numerator and denominator by j
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(((3-4j)(3j-2))/-j) * (j/j) =
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Note the denominator becomes -j^2 = -(-1) = 1
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(((3-4j)(-2+3j)) * j =
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perform complex number multiplication
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(-6 +9j +8j -12j^2) * j =
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(-6 +9j +8j +12) * j =
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(6 +17j) * j =
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6j -17 =
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-17 +6j
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modulus = square root( (-17)^2 + 6^2 ) = 18.0278
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