Complex numbers- locus- hard (1 Viewer)

[lsdpoon1337]

1.If Re(z^2 + z) is a constant, find the locus of z
2.If Re (z) is a constant find the locus of z^2 + z
3.p^2= p +1 find p^(-5) as linear polynomial in terms of p

thanks

 

[shaon0]

1.If Re(z^2 + z) is a constant, find the locus of z
2.If Re (z) is a constant find the locus of z^2 + z
3.p^2= p +1 find p^(-5) as linear polynomial in terms of p

thanks

Let z=x+iy.
1) Re(x^2+3ixy-y^2+x)
=> x^2-y^2+x=a where a is a constant.
=> (x+1/2)^2-y^2=a+0.25
Which is a hyperbola.

2) Re(z)=x=a where a is a constant.
Thus, z^2+z
= x^2+x+3ixy-y^2
= a^2+a+3iay-y^2
Let C be a constant.
= C+3iay-y^2

3) p^7=(p+1)^3.5
p^(-5)=p^2/p^7
=(p+1)/(p+1)^(3.5)
= (p+1)^(-5/2)
You might have to square both sides after this.

 

[kwabon]

3) p^7=(p+1)^3.5
p^(-5)=p^2/p^7
=(p+1)/(p+1)^(3.5)
= (p+1)^(-5/2)
You might have to square both sides after this.

sorry mate , ummm could be please expand on that ..... and the question said express p as a polynomial, and i dont see how (p+1)^2.5 is a polynomial, if that sentence in bold is gonna correct it could u please show me how

thanks mate

 

[Iruka]

p^2=p+1.

Since p isn't zero, we may divide through to get p=1+p^-1

Rearranging this, we have p^-1 = p-1

Squaring, we obtain p^-2 = p^2-2p+1, but since p^2=p+1, we simplify this to p^-2 = -p+2. Square again, and then multiply by p^-1, using the equation p^2=p+1 to simplify at each step. Then you can express p^-5 as a linear polynomial in p.

 

[lsdpoon1337]

ah yes, that all makes sense

thanks

 

[kwabon]

^^ sure does
thanks