anonymous
  • anonymous
Consider the following complex numbers: 1 + i 1.5 + i 1.5 − i 1 + 0.5 i −1 + i 1 − 0.5 i How many are a fourth root of −4?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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cwrw238
  • cwrw238
i know 1 + i is
anonymous
  • anonymous
any others??
jim_thompson5910
  • jim_thompson5910
Finding the fourth root of -4 is the same as solving the equation x^4 = -4 which becomes x^4 + 4 =0 So it's a polynomial of degree 4. So there will be 4 complex roots. Luckily, complex roots come in conjugate pairs. So if we know that 1+i is a root (from cwrw238), then 1-i is also a root. This means that x = 1+i or x = 1-i x - (1+i) = 0 or x-(1-i) = 0 (x - (1+i))(x-(1-i)) = 0 ( (x-1) - i)( (x-1) +i) = 0 (x-1)^2 - i^2 = 0 (x-1)^2 + 1 = 0 x^2 - 2x+1+1 = 0 x^2 - 2x + 2=0 So this means that x^2 - 2x + 2 is a factor of x^4 + 4 To find the other factor, use polynomial long division x^2 + 2x + 2 ------------------------------ x^2 - 2x + 2 | x^4 + 0x^3 + 0x^2 + 0x + 4 x^4 - 2x^3 + 2x^2 ---------------- 2x^3 - 2x^2 2x^3 - 4x^2 + 4x + 4 ----------------- 2x^2 - 4x + 4 2x^2 - 4x + 4 -------------- 0 This does two things really: First is the remainder of 0 confirms that x^2-2x+2 is really a factor of x^4 + 4. Second is that the quotient x^2+2x+2 shows us that x^4 + 4 = (x^2-2x+2)(x^2+2x+2) From here, all you need to do is solve x^2+2x+2 = 0 to find the other two roots. Use the quadratic formula to solve x = (-b+-sqrt(b^2-4ac))/(2a) x = (-(2)+-sqrt((2)^2-4(1)(2)))/(2(1)) x = (-2+-sqrt(4-(8)))/(2) x = (-2+-sqrt(-4))/2 x = (-2+sqrt(-4))/2 or x = (-2-sqrt(-4))/2 x = (-2+2*i)/2 or x = (-2-2*i)/2 x = -1 + i or x = -1-i So in total, there are 4 roots and they are 1+i, 1-i, -1+i, -1-i

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