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Question

For c0, find number of complex numbers z which satisfy the equation |z|22iz+2c(1+i)=0.

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Solution

Put z=x+iy.
Then equation takes the form
x2+y22i(x+iy)+2c(1+i)=0
(x2+y2+2y+2c)+2i(cx)=0
Equating real and imaginary parts to zero, we get
x2y2+2y+2c=0,2c2x=0x=c ...(1)
These give x=c and for y on putting for x, we get
y2+2y+c2+2c=0
y=2±4(1c22c)2 ...(2)
where 1c22c0 for real yor 2c2+2c+1 or (c+1)2(2)20
2(c+1)2
21c21
Since c0 we contract the above interval to. 0c21
I. c=0(x,y)=(0,1±1) or (0,0),(0,2)
II. 0<c<(21).
There are two solutions x,y given by (1) and (2).
III. c=21i.e.1c22c=0, then from (1) and (2) there is only one solution from (1) and (2), (x,y)=(c,1)
IV. For c>21,Δ<0 and hence there is no solution.

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