Complex numbers, polynomials (1 Viewer)

YBK

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hey, any help on this question would be greatly appreciated
I can do the first part, but not the second.


Let F(z) = z^8 - 7/3 z^4 + 1
The complex number a is a root of F(z) = 0

i) Show that Ia and 1/a are also roots of F(z) = 0

a^8 - 7/3 a^4 + 1 = 0

(Ia)^8 - 7/3 (Ia)^4 + 1
= a^8 - 7/3a^4 + 1
=0

(1/a)^8 - 7/3 (1/a)^4 + 1
= {1 + 7/3 a^4 + a^8}/a^8
= 0


ii) Find all the roots of P(z) = 0


Can't get far with this part.


Thanks :)
 

KeypadSDM

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As it's a quadratic in quartic terms, the value of the function is invariate when you input a and i * a

I.e.:

F(iz) = (iz)8 - 7/3 * (iz)4 + 1 = z8 - 7/3 * z4 + 1 = F(z), smiliarly,
F(-z) = F(z)

And, as the coefficients are real, we have:

If a is a root (of the form x + iy), then so is:
a, -a, ia, -ia
and
b, -b, ib, -ib

where b = x - iy

So let's just solve the quadratic part first:
F(z) = z8 - 7/3 * z4 + 1

2 * z4 = 7/3 +Sqrt[49/9 - 4 * 1 * 1] = 7/3 + Sqrt[13]/3 = 1/3(7 + Sqrt[13])

Then proceed from there.
 

YBK

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KeypadSDM said:
So let's just solve the quadratic part first:
F(z) = z8 - 7/3 * z4 + 1

2 * z4 = 7/3 +Sqrt[49/9 - 4 * 1 * 1] = 7/3 + Sqrt[13]/3 = 1/3(7 + Sqrt[13])

Then proceed from there.
Isn't a quadratic equation one with leading coeficient 2?

Are you sure we can use the quadratic formula on that?
 

KeypadSDM

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YBK said:
Isn't a quadratic equation one with leading coeficient 2?

Are you sure we can use the quadratic formula on that?
It's just algebra, this method is always correct.

The method is derived from completion of the square, which is still valid if you do it with z^4. In fact this is always valid as the complex plane comprises all solutions to all complex polynomials of all orders.
 

YBK

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thanks keypad

but how do you go on from there?


The solution to this problem from the book Pheonix Questions by topic is wrong...
 

KeypadSDM

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Well just take the real positive fourth root of either of the solutions to the quadratic [a say] then the 8 roots are:

a, ia, -a, -ia
1/a, i/a, -1/a, -i/a

And all of those 8 answers WILL be different.
 

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