Which is the value for |z|^2? The complex number z is the solution of the equation |z|+z = 1 + (2+i)/sqrt3.

giorgiana1976 | Student

First, we'll set the complex number as z=a+b*i.

Also, we know that |z|=sqrt(a^2+b^2)

Now, we'll substitute z and |z| into the given relation.

|z|+z = sqrt(a^2+b^2) + a+b*i

sqrt(a^2+b^2) + a+b*i = 1 + (2+i)/sqrt3

For the identity to be true, the real parts from both sides have to be equal and the imaginary parts from the both sides, have to be equal, too.

For finding the real part and the imaginary part from the right side, we have to open the brackets.

1 + (2+i)/sqrt3 = 1 + (2/sqrt3) + (1/sqrt3)*i

So, we'll have:

sqrt(a^2+b^2) + a = 1 + (2/sqrt3)

and

b = 1/sqrt3

sqrt(a^2 + 1/3) = 1 + (2/sqrt3) - a

a^2 + 1/3 = [1 + (2/sqrt3)]^2 - 2a*[1 + (2/sqrt3)] + a^2

After reducing similar terms, we'll get:

2a + 4a/sqrt3 = 1 + 4/sqrt3 + 4/3 - 1/3

We'll factorize and we'll get:

a(2 + 4/sqrt3)  = 2 + 4/sqrt3

a=1

So, |z|^2 = a^2 + b^2

|z|^2 = 1 + 1/3

|z|^2 = (3+1)/3

|z|^2 = 4/3

neela | Student

|z|+z = 1+(2+i)/sqrt3. To find |z|^2.

Since z  is a coplex number we can assume it tobe x+y*i, where x and y are real and x is the real part and y*i is the imaginary part and i =  sqrt (-1).

Also: |z| = |x+y*i| = sqrt(x^2+y^2) by defonition.

So,  LHS = |z|+z = |x+y*i| +x+y*i

= sqrt(x^2+y^2)+x+y*i. = {sqrt(x^2+y^2)+x}  + y*i

The RHS of the given equation:2+(2+i)/sqrt3 = {1+2/sqrt3} +i/sqrt3.

Now we equate the real parts of the LHS and the RHS and get:

sqrt(x^2+y^2)+x = 1+2/sqrt3. Or

sqrt(x^2+y^2) = 1+2/sqrt3 -x..............(1)

Equating the imaginary parts on  LHS and RHS,we get:

yi=i/sqrt3. Or

y = 1/sqrt3 = sqrt3/3  with the  rationalised denominator.Substituting this value in (1) we get:

sqrt(x^2+i^2/3) -x = 1+2/sqrt3. Or

sqrt(x^2-1/3) = 1+2/sqrt3 -x. Squaring both sides, we get:

x^2-1/3 = 1+4/sqrt3+4/3 +x^2- 2x(1+2/sqrt3). Or

2x(1+2/sqrt3) = 8/3 +4/sqrt3. Or

x = (1/3){4+2/sqrt3)/(1+2/sqrt3) = (1/3){4sqrt3+2}/{sqrt3+2}

=(2/3) (2sqrt3 +1) (2-sqrt3)/ (2^2-3)

=(2/3) {4sqrt3-6+2-sqrt3}/1

=(2/3){3sqrt3-4}