Question 205743: find all solutions of the equation z^2 = -6 + 8i
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! find all solutions of the equation z^2 = -6 + 8i
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z = sqrt(-6 + 8i)
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Convert to trig form:
r = sqrt(6^2+8^2) = 10
theta = arctan(8/-6) = -53.13;
But z^2 is in the 2nd quadrant so theta = -53.13+180 = 126.87
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z = sqrt(10)(cis(126.87+360n)/2)
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let n = 0, then z = sqrt(10)(cis(63.435))
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let n = 1, then z = sqrt(10)(cis(243.435)
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In Rectangular form
For n=0
x = sqrt(10)cos(63.435) = 1.414 or sqrt(2)
y = sqrt(10)sin(63.435) = 2.828
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For n = 1
x = sqrt(10)cos(243.435) = -sqrt(2)
y = sqrt(10)sin(243.435) = -2.828...
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Cheers,
Stan H.
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