Find the number of different ordered quadruples (a,b,c,d) of complex numbers such that
a^2=1
b^3=1
c^4=1
d^6=1
a+b+c+d=0
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solve a^2=1, b^3=1, c^4=1, d^6=1, a+b+c+d=0
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There are 3 solutions with real numbers and 4 solutions with complex numbers
To find the number of different ordered quadruples (a,b,c,d) of complex numbers satisfying the given conditions, we need to analyze each condition and use the information to narrow down the possibilities.
Condition 1: a^2 = 1
This condition means that a can take on two possible values: a = 1 or a = -1.
Condition 2: b^3 = 1
Similarly, this condition means that b can take on three possible values: b = 1, b = e^(2πi/3), or b = e^(4πi/3), where e is the base of the natural logarithm (approximately 2.71828) and i is the imaginary unit.
Condition 3: c^4 = 1
This condition means that c can take on four possible values: c = 1, c = i, c = -1, or c = -i, where i is the imaginary unit.
Condition 4: d^6 = 1
Finally, this condition means that d can take on six possible values: d = 1, d = e^(πi/3), d = e^(2πi/3), d = e^(πi), d = e^(4πi/3), or d = e^(5πi/3), where e is the base of the natural logarithm (approximately 2.71828) and i is the imaginary unit.
Now, let's consider the condition a + b + c + d = 0.
Case 1: a = 1
If a = 1, then b + c + d = -1. In this case, there are (3)(4)(6) = 72 different possibilities for (b, c, d) since b, c, and d can each take on different values.
Case 2: a = -1
If a = -1, then b + c + d = 1. Similarly, there are (3)(4)(6) = 72 different possibilities for (b, c, d) since b, c, and d can each take on different values.
Therefore, the total number of different ordered quadruples (a, b, c, d) is 72 + 72 = 144.