Calculate the module of the complex number z+2z'=3+i.

Expert Answers
hala718 eNotes educator| Certified Educator

z+ 2z' = 3 +i

Let z = a+bi

==> z= a-bi

==> z + 2z' = a+bi  + 2(a-bi)

                   = a+bi + 2a - 2bi

                    = 3a - bi

==> 3a - bi = 3 +i

==> a = 1  ==> b = -1

==> z = 1 - i

lzl = sqrt(a^2 + b^20

      = sqrt(1 +1) = sqrt2

==> lzl = sqrt2

giorgiana1976 | Student

The module of a complex number is:

|z| = sqrt [(Re z)^2 + (Im z)^2]

If z has is written algebraically

z = a + b*i, then the real part = Re z = a and the imaginary part is Imz = b.

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

The conjugate of z is z' = a - b*i

Now, we'll re-write the given expression:

z+2z'= a + b*i + 2a - 2 b*i

We'll combine the real parts and the imaginary parts:

(a+2a)+i*(b-2b)=3+i (from enunciation)

The real part from the left side has to be equal to the real part from the right side.

a+2a=3, a=1

-b*i= i

b=-1

The complex number z is:

z=1-i

|z| = sqrt(a^2 + b^2), where a = 1 and b = -1

|z| = sqrt(1 + 1)

|z| = sqrt 2