Question

Solve for z, i.e, find all complex numbers z which satisfy $∣z∣^2−2iz+2c(1+i)$=0, where c is real.

Answer 1

Hint: We need to form equations from the question, put value of z=x+iy then form equation after that solve the quadratic equation to get value of c.

Complete step-by-step answer:
Put $$z = x + iy.$$
Then the given equation reduces to
$${x^2} + {y^2} - 2i(x + iy) + 2c(1 + i) = 0$$
Or ($$({x^2} + {y^2} + 2y + 2c) + 2i(a - x) = 0$$
Equating the real and imaginary part to zero, we get
$${x^2} + {y^2} + 2y + 2c = 0,2c - 2x = 0$$
This gives for x=c and for y we have
$${c^2} + {y^2} + 2y + 2c = 0\;or\;{y^2} + 2y + ({c^2} + 2c) = 0\;\;$$...(1)
Since we seek real value of y, the discriminant of the equation (1)
must be non-negative, that is
△=$$4 - 4({c^2} + 2c) \geqslant 0 \Rightarrow 1 - {c^2} - 2c \geqslant 0$$
For the real value of c, we get
$$y = \dfrac{{ - 2 \pm \sqrt {4(1 - {c^2} - 2c)} }}{2} = - 1 \pm \sqrt {1 - {c^2} - 2c} $$
Hence for △≥0, the original equation has two roots
$$\eqalign{
  & {z_1} = c + ( - 1 + \sqrt {1 - {c^2} + 2c} )i \cr
  & {z_2} = c + ( - 1 - \sqrt {1 - {c^2} + 2c} )i \cr} $$
For △=0, we have $${z_1} = {z_2}$$that is,
there is only one solution in this case.
For △<0 the equation has no roots
It remains to indicate the range of c over which there are solutions.
So c must satisfy the inequality.
$$\eqalign{
  & 1 - {c^2} - 2c \geqslant 0 \Rightarrow {c^2} + 2c - 1 \leqslant 0 \cr
  & \Rightarrow {(c + 1)^2} - 2 \leqslant 0{(c - 1)^2} \leqslant 2 \cr
  & \Rightarrow - \sqrt 2 \leqslant c + 1 \leqslant \sqrt 2 \cr
  & \Rightarrow - 1 - \sqrt 2 \leqslant c \leqslant - 1 + \sqrt 2 \cr} $$

Note: The set of complex numbers, denoted by C, includes the set of real numbers (R) and the set of pure imaginary numbers.