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

```
9 months ago

2056 Points
```							let z= x+iyso x+iy + i + (root 2) | z +1|= 0x + (root 2) | z +1| + i(y+1)= 0equating real and imaginary parts to 0,y+1 = 0 and x + (root 2) | z +1| = 0so y= – 1and – 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= 0x= – 2so, z= – 2 – ikindly approve :))
```
9 months ago
Vikas TU
12141 Points
```							Dear student Please check out the study materials for more such exampleshttps://www.askiitians.com/iit-jee-algebra/complex-numbers/Good Luck
```
9 months ago
Srutarshi Tripathi
38 Points
```							Since , z + \/2 I z+1 I + i = 0the middle term is purely real. So, Im(z+i)=0 or Im(z) + 1 =0 or, Im(z) = -1Now, let z= x – iHence , the equation reduces to :x + \/(2( (x+1)2 + 1) ) =0or, x2 = 2(x+1)2 + 2or, x2 = 2x2 + 4x + 4or, x2 + 4x + 4 =0or, x = – 2So, z = – 2 – i (ANSWER)
```
6 days ago
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