# Thread: locus of complex no. question no. 2

1. ## locus of complex no. question no. 2

p294 q16 b c
question:
it is given that the equation
$\displaystyle z^2-2z+k =0$(k is real ) --------- (*)
has no real roots.
a) find the range of k
b) find the quadratic equation whose roots are the cubes of the roots of (*) and show that the discriminant of this equation is $\displaystyle 4(1-k)(4-k)^2$
if this equation has real roots , deduce the value of k.
c) find, in terms of k , the squares of the roots of (*), expressing the answers in the form x+iy where x and y are real.
as k varies, find the equation of the locus of the points in the Argand plane representing the squares of the roots of (*).

i have problems in b) and c)
b) $\displaystyle 4(1-k)(4-k)^2 \ge 0$
$\displaystyle (k-1)(k-4)^2 \le 0$
$\displaystyle k \le 1$ or$\displaystyle k = 4$
But the answer says it is k = 4. i don't understand.
c)i get the two points $\displaystyle 2 - k + 2\sqrt{1-k}$and $\displaystyle 2 - k - 2\sqrt{1-k}$ ie $\displaystyle -2 + 2 \sqrt 3 i$and $\displaystyle -2 - 2 \sqrt 3 i$
but i don't know how to represent the locus of them as k varies. it seems that i have problems in the idea of locus.
thanks!

2. Originally Posted by afeasfaerw23231233
b) $\displaystyle 4(1-k)(4-k)^2 \ge 0$
$\displaystyle (k-1)(k-4)^2 \le 0$
$\displaystyle k \le 1$ or$\displaystyle k = 4$
But the answer says it is k = 4. i don't understand.
Isn't that because (*) is supposed to have no real roots ?
And you should have found in a) that k has to be superior (never equal) to 1. So that eliminates solution $\displaystyle k \le 1$.

But I don't see where your working comes from... Have you dealt with the first part of the question ?