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z + ( root 2) | z +1| +i=0 z is a complex number then z is equal to

z + (root 2) | z +1|+i=0
z is a complex number then z is equal to 
 

Grade:12

3 Answers

Aditya Gupta
2081 Points
4 years ago
let z= x+iy
so x+iy + i + (root 2) | z +1|= 0
x + (root 2) | z +1| + i(y+1)= 0
equating real and imaginary parts to 0,
y+1 = 0 and x + (root 2) | z +1| = 0
so y= – 1
and – x= (root 2) | z +1|= (root 2) | x+1 + iy|
x^2= 2[(x+1)^2 + y^2]= 2[(x+1)^2 + 1]= 2(x^2+2x+2)
or x^2= 2x^2+4x+4
(x+2)^2= 0
x= – 2
so, z= – 2 – i
kindly approve :))
Vikas TU
14149 Points
4 years ago
Dear student 
Please check out the study materials for more such examples
Good Luck 
Srutarshi Tripathi
38 Points
3 years ago
Since , z + \/2 I z+1 I + i = 0
the middle term is purely real. So, Im(z+i)=0 or Im(z) + 1 =0 
or, Im(z) = -1
Now, let z= x – i
Hence , the equation reduces to :
x + \/(2( (x+1)2 + 1) ) =0
or, x2 = 2(x+1)2 + 2
or, x2 = 2x2 + 4x + 4
or, x2 + 4x + 4 =0
or, x = – 2
So, z = – 2 – i (ANSWER)

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