Find the three roots of the equation z^3 = 8i giving your answers in the form x+iy? - find your roots
I can not do this question at all!
Saturday, January 9, 2010
Find Your Roots Find The Three Roots Of The Equation Z^3 = 8i Giving Your Answers In The Form X+iy?
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3 comments:
Using the formula a ^ 3 + b ^ 3 = (a + b) (a ^ 2 - ab + b ^ 2)
z ^ 3 = 8i
=> Z ^ 3 - 8i = 0
=> (Z) ^ 3 + (2i) ^ 3 = (z + 2i) (z ^ 2 - 2ZI - 4) = 0
if z + 2i = 0 or
z ^ 2 - 2ZI - 4 = 0
where z + 2i = 0
z =- 2i
where z ^ 2-2ZI - 4 = 0
z = [2i + / - sqrt (-4 16)] / 2
z = [2i + / - sqrt (12)] / 2
z = [2i + / - 2sqrt (3)] / 2
z = i + sqrt (3) oi - sqrt (3)
Thus, three roots 2i [are i + sqrt (3)], [i - sqrt (3)]
One obvious result is 0-2i. The roots are much more difficult, an equal number of intensive use
(x + iy) ^ 3 = x ^ 3 + 3x ^ 2iy - 3xy ^ 2 - i ^ 3
If this equation is equal 8i, then Defenition an equal number of complex real and imaginary part must be proportional and equal:
x ^ 3 - 3xy ^ 2 = 0
3x ^ 2 (y) - y ^ 3 = 8
and the resolution for the replacement of x is in the second equation
x = sqrt [(8 + y ^ 3) / 3y]
to solve and for the first
sqrt [(8 + y ^ 3) / 3y] ^ 3-3sqrt [(8 + y ^ 3) / 3y] y ^ ... = 0
sqrt [(8 + y ^ 3) / 3y] ^ 3 = 3sqrt [(8y ^ 3 + y ^ 6) / 3]
(8 + y ^ 3) ^ 3 / (27y ^ 3) = 9 [(8y ^ 3 + y ^ 6) / 3]
(8 + y ^ 3) ^ 3 = 81 and ^ 3 (8y ^ 3 + y ^ 6)
y 8 y 9 ^ ^ ^ 6 192 3 512 = 648y ^ 6 and 81 ^ 9
80Y-9-640y ^ ^ 6 192 3 512 = 0
temporarily the use of V ^ 3, the equation of a cubic
80V-640V-3 ^ ^ 2 192 V 512 = 0
Suppose z = x + iy
Then z ^ 2 = x ^ 2 - y ^ 2 + y 2xyi z * i = - y ^ 2 + xi
z ^ 3 = 8i = - (2i) ^ 3
z ^ 3 + (2i) ^ 3 = 0
(z 2 i) * [z ^ 2 + (2i) ^ 2 - 2ZI] = 0
Therefore, z + 2i = 0, ie x = 0 and y = -2
Oz ^ 2 + (2i) ^ 2 - 2ZI = 0
x ^ 2 - y ^ 2 +2 xy - 4i - 2 (- y ^ 2 + xi) = 0
x ^ 2 + y ^ 2 + (2xy-2x -4) i = 0
if and only if:
x ^ 2 + y ^ 2 = 0 and 2xy-x-4 = 0 (2)
Since there is no real root of (2)
Then the given equation is a root z =- 2i.
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