Roots of a squared polynomial ( complex numbers)

Axe199
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Homework Statement



problem in a pic attached

Homework Equations





The Attempt at a Solution


i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i don't know what he wants from me
 

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Axe199 said:

Homework Statement



problem in a pic attached

Homework Equations





The Attempt at a Solution


i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i don't know what he wants from me

He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.
 
Ray Vickson said:
He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.

why can't i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?
 
Axe199 said:
why can't i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?

(z + 2)2 is not a factor of p(z2). It is a factor of (p(z))2, however, a completely different polynomial.
The polynomial is not the object being squared. p(z2) only means that the variable z is being replaced with the variable z2. So, as a result, we know that z2 + 2 is a factor.
 
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slider142 said:
(z + 2)2 is not a factor of p(z2). It is a factor of (p(z))2, however, a completely different polynomial.
The polynomial is not the object being squared. p(z2) only means that the variable z is being replaced with the variable z2. So, as a result, we know that z2 + 2 is a factor.
okay...thanks very much, tht was a tricky one, or maybe i am stupid :D
either way , thanks , i got it
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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