Complex numbers problem...?

Question

If z is a complex number of the form a+ib, what is the set of z which satisfies this equation:

z + (conjugate of z) = (modulus of z)^2?

In other words, (a + ib) + (a - ib) = a^2 + b^2?

Answer 1

If z=a+ib, then we have:
2a=a²+b²
a²-2a+b²=0
This is a quadratic equation in a, with solutions:

a=(2±√(4-4b²)/2
a=1±√(1-b²)

Note that since both a and b are real numbers, this implies that 0≤a≤2 and -1≤b≤1.

Answer 2

(a + ib) + (a - ib) = a^2 + b^2
it means
2a =a^2 +b^2,
a^2 -2a +b^2 =0,
so it is a quadratic equation:
a = ( 2 +- sqrt( 4 -4(b^2)) /2
= ( 2 +- sqrt(4(1-b^2)) ) /2
= 2 +- sqrt( 1-b^2)
now, this is a solution if
1-b^2 >= 0
1>= b^2
b^2 <= 1 and this happens if -1<= b <=1

so the set of complex numbers is given by:
z= 2 +- sqrt( 1-b^2) + ib, where -1<= b <=1
'

Answer 3

a + ib +a -ib = a^2 + b^2
a^2 - 2a +b^2 = 0
ab sab main hi bataun
quadraticly solve karo aur a aur b mein relation nikalo

Answer 4

in view that x = -2+2i is one root, different root is -2 – 2i. made of those 2 roots = 8. different 2 roots are {-6±?(-sixty 4)}/10 = (-6 ± 8i)/10 = -3/5 ± 4i/5. made of those 2 roots = 5. => f = a million, a = 5, g = 4, e = 40. comparing coeffs of powers of x, b = 5g + 6f = 26, c = 40 + 5f + 6g = sixty 9 and d = 40 8 + 5g = sixty 8.

Answer 5

0 = a^2 + b^2 - 2a = (a-1)^2 - 1 + b^2
(a-1)^2 + b^2 = 1

So it's a circle of radius 1 centered at (1, 0) or at z = 1.

Answer 6

As far as I can see, both Fred and Pascal are right -- it's just a question of whether you express it in terms of a or b. I don't know why 2 people gave Fred's answer a minus rating. It's actually a slightly simpler solution that Pascal's, but Pascal did give the restrictions on the values of a and b. I don't know which one you should give the points to.

Answer 7

2a = a^2 + b^2
if b = zero, a= 0 or a = 2

u can express a in terms of b and get the set:
z = a +- i(sqrt(2a-a^2))

Answer 8

b must be zero

for the real part:

2a=a^2 , go ahead, solve for it.

When

so, a must equal 2, b must be zero.