# 2z + z' = 3 + 2i find lzl

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### 3 Answers

2z + z' = 3 + 2i

let z = a+bi

==> z' = a-bi

Now substitute:

==> 2(a+bi) + (a-bi) = 3 + 2i

==> 2a + 2bi + a - bi = 3 + 2i

Group like terms:

==> 3a + bi = 3 + 2i

==> 3a = 3 ==> **a= 1**

==>** b = 2 **

**==> z= 1+2i**

We know that:

lzl = sqrt(a^2 + b^2)

= sqrt(1 + 4) = sqrt4

**==> lzl = sqrt5**

The module of a complex number z = a + i*b is the positive square root of the sum of the squares of the real part, a, and imaginary part, b.

The conjugate of a complex number z is z' = a - b*i

We'll calculate the sum from the left side:

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

We'll remove the brackets:

2a + 2b*i + a - b*i

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

2z + z' = 3a + b*i

The real part and the imaginary part of the complex number from the left side have to be equal with the real part and the imaginary part of the complex number form the right side:

In our case, the complex number from the right side is z = 3 + 2i

3a + b*i = 3 + 2i

We'll identify the real part and the imaginary part:

Real part - Re(z) = a:

3a = 3

We'll divide by 3:

a = 1

a = 3

Imaginary part - Im(z) = b

b = 2

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

|z| = sqrt (1^2 + 2^2)

**|z| = sqrt (5)**

2z+z' = 2+2i. To find |z|.

We presume z = x+yi, where x and y are real and i = sqrt(-1).

Then z' conjugate of z = conjugate of (x+yi) = x-yi.

Now the given equation becomes:

2(x+yi )+(x-yi) = 3+2i.

2x+2yi+x - yi = 3+2i

3x+yi = 3+2i.

Equate real parts on both sides and equate imaginary parts on both sides:

3x= 3. So x = 3/3 = 1.

y = 2.

So z = x+yi = 1+2i.

So |z| = |x+yi| = sqrt(x^2+y^2) =sqrt(1+2^2) = sqrt5.