answers can be 54, 61, 49, 52 and none
2006-12-03 09:27:03 · 6 answers · asked by scoota 1 in Science & Mathematics ➔ Mathematics
What is your question?
I will help you if you tell me what you are solving for.
Is (4 + 6i) equal to something?
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Rhsaunder is correct.
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If we're directed to multiply by the complex number's conjigate. . .
Then, the product is 52.
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GIVEN:
(4 + 6i)
EVALUATE:
by multiplying by its conjigate.
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REMEMBER:
A conjigate is identical, except for the sum or difference.
If the factor is a sum of two numbers. . .
Then, it's conjigate will be the difference of those same two numbers.
The conjigate of (4 + 6i) is (4 - 6i).
To get the product, we MULTIPLY the complex number with its conjigate.
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SOLUTION:
(4 + 6i)(4 - 6i)
16 -24i + 24i - 36i²
16 + 0 - 36i²
16 - 36i²
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REMEMBER:
i² = -1
SUBSTITUTE the known value of i² using (PARENTHESIS).
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16 - 36i²
16 - 36(-1)
16 + 36
SIMPLIFY by combining "like" terms.
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FINAL ANSWER:
52
Hope that was the question!
And this is what you were looking for. ♥
2006-12-03 09:31:28 · answer #1 · answered by LovesMath 3 · 1⤊ 0⤋
You don't show a question.
4 + 61 is a complex number.
54, 61, 49 and 52 are integers.
MAYBE you have to calculate (4+6i)(4-6i).
It is 16 + 36 = 52
4-6i and 4+61 have been 'conjugated'.
MAYBE you need the modulus of 4 + 6i
It is 16 + 36 = 52 as well
Th
2006-12-03 09:33:42 · answer #2 · answered by Thermo 6 · 1⤊ 0⤋
The question is incomplete. The complex number 4+6i is equal to none of those. However, the product of that number and its complex conjugate is 52, so that is probably the answer that is wanted.
2006-12-03 09:30:05 · answer #3 · answered by Anonymous · 1⤊ 0⤋
i am assuming that you are adding vectors...
if this is the case the it is simply solved by using a right traingle
4^2 + 6^2 = 52. take the square root of this and you would get the answer. based on your answers above it would be none i guess.
2006-12-03 09:35:06 · answer #4 · answered by Mike 2 · 1⤊ 0⤋
|4+6i| = â(4² + 6²)
= â(52)
OR |4+6i|² = 52
I am unsure what you are seeking but this was one way I could see to get one of the answers you had
2006-12-03 09:34:08 · answer #5 · answered by Wal C 6 · 0⤊ 2⤋
that's not a complete question.
Is i supposed to be the imaginary number sqrt(-1), or a variable?
2006-12-03 09:30:45 · answer #6 · answered by Jim Burnell 6 · 0⤊ 0⤋