SOLUTION: 4th root of -256

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Question 987077: 4th root of -256
Answer by ikleyn(52748) About Me  (Show Source):
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Notice that     -256 = 256(cos(180°) + i*sin(180°))   and   256 = 4%5E4.

The roots of degree  4  of  -256  are  4  complex numbers

1)  4(cos(45°) + i*sin(45°)) = 4%28sqrt%282%29%2F2+%2B+i%2A%28sqrt%282%29%2F2%29%29 = 2sqrt%282%29+%2B+2sqrt%282%29%2Ai;                                                                         (Notice that 45° = 180°/4)

2)  4(cos(45°+90°) + i*sin(45°+90°)) = 4(cos(135°) + i*sin(135°)) = -4%2Asqrt%282%29%2F2+%2B+%28%284%2Asqrt%282%29%29%2F2%29%2Ai = -2sqrt%282%29+%2B+2sqrt%282%29%2Ai;         (Notice that 90° = 360°/4)

3)  4(cos(45°+180°) + i*sin(45°+180°)) = 4(cos(225°) + i*sin(225°)) = -4%2Asqrt%282%29%2F2+-+%284%2Asqrt%282%29%2F2%29%2Ai = -2sqrt%282%29+-+2sqrt%282%29%2Ai;

4)  4(cos(45°+270°) + i*sin(45°+270°)) = 4(cos(315°) + i*sin(315°)) = 4%2Asqrt%282%29%2F2+-+%284%2Asqrt%282%29%2F2%29%2Ai = 2sqrt%282%29+-+2sqrt%282%29%2Ai.

If you want to see my lessons on complex numbers in this site,  look in this
    - REVIEW of lessons on complex numbers
and especially in this one
    - How to take a root of a complex number.