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Algebra
teja's Avatar
teja
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9 Apr 2009 18:03:19 IST
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please answer me no. of solutions of z^6+z+1=0?
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please answer me no. of solutions of z^6+z+1=0?


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AS GOOD AS ROSE's Avatar

AS GOOD AS ROSE
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9 Apr 2009 18:04:12 IST
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is the answer four
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AS GOOD AS ROSE's Avatar

AS GOOD AS ROSE
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9 Apr 2009 18:04:18 IST
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is the answer four
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chirag gagrani
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9 Apr 2009 18:55:07 IST
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No. of solutions will be 6,whether real or complex,according to degree of eq.
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Bipin Dubey
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9 Apr 2009 23:57:31 IST
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f(z) = z6 + z + 1

 

f '(z) = 6z5 + 1 = 0  gives  z = (-1/6)1/5

 

f ''(z) = 30z44 > 0

 

so the point z = (-1/6)1/5 is a minima.

 

Since the curve has only minima and no maxima it cuts the x-axis at only two points.

 

So there are are two real roots for the given equation and four imaginary roots.

 

 
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Siddhartha Gupta's Avatar

Siddhartha Gupta
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10 Apr 2009 13:18:10 IST
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But if it has one minima , how can you be sure it cuts the x-axis in two pts? Can't it cut at one pt?
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Bipin Dubey
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24 Apr 2009 12:49:06 IST
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One more step which I missed,

 

f((-1/6)1/5) = (1/6)6/5 + (-1/6)1/5 + 1 > 0

 

so f(z) lies above the axis for all values, since it never cuts or touches the x-axis it has no real solution and 6 imaginary solution.

 

 
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