# 1)  FIND THE NUMBER OF ROOTS BETWEEN lZl=1 AND lZl=2, OF Z7(POWER=7) + 4Z2(POWER=2) +11=02) A,B,C,D ARE COMPLEX NUMBERS..GIVEN THAT NUMBERS (A-D/B-C) AND (B-D/C-A) ARE PURELY IMAGINARY..PROVE THAT C-D/A-B IS ALSO PURELY IMAGINARY.

mycroft holmes
272 Points
14 years ago
2. Its easy to see that a,b,c cannot be collinear. Remember that if z1/z2 is purely imaginary then the radius vectors 0Z1 and OZ2 are perpendicular to each other Then we can draw triangle ABC with a,b, and c as the vertices. From the statement of the problem, the complex number is the intersection of altitudes from vertex a and b. Since the orthocentre is unique for any triangle, d also lies on the altitude dropped from c. i.e. c-d is perpendicular to a-b or c-d/a-b is purely imaginary
mycroft holmes
272 Points
14 years ago
We will prove that all roots will lie in this region. Suppose |z|<1, then z^7 + 4z^2+11 = 0 implies 11 = |z^7+4z^2| <= |z|^7 + 4 |z|^2 <=5 a contradiction On the other hand if |z|>=2, 11 = |z^7+4z^2| >= |z|^7 - 4 |z|^2 Now f(x) = x^7 - 4x^2 is an increasing function for x>=2, and f(2)>11. So again we have a contradiction. So all the roots got to lie in the region 1<|z|<2.