Solveeit Logo
Login

Question

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

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

Explanation

Solution

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.z = x + iy.
Then the given equation reduces to
x2+y2−2i(x+iy)+2c(1+i)=0{x^2} + {y^2} - 2i(x + iy) + 2c(1 + i) = 0
Or ((x2+y2+2y+2c)+2i(a−x)=0({x^2} + {y^2} + 2y + 2c) + 2i(a - x) = 0
Equating the real and imaginary part to zero, we get
x2+y2+2y+2c=0,2c−2x=0{x^2} + {y^2} + 2y + 2c = 0,2c - 2x = 0
This gives for x=c and for y we have
c2+y2+2y+2c=0  or  y2+2y+(c2+2c)=0    {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(c2+2c)⩾0⇒1−c2−2c⩾04 - 4({c^2} + 2c) \geqslant 0 \Rightarrow 1 - {c^2} - 2c \geqslant 0
For the real value of c, we get
y=−2±4(1−c2−2c)2=−1±1−c2−2cy = \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

& {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.