# Thread: Absolute Value of Complex Numbers

1. ## Absolute Value of Complex Numbers

I've been staring at this problem for quite some time...

Let z belong in C such that |z| = 1. Compute |1+z|^2 + |1-z|^2

If it was like |z + w|^2 where w was also a complex number, then that would be a lot like multiplying vectors using the dot product, correct? So it is much simpler where w is a real number? I'm not sure if I understand how to go about this...

- Nicole

2. Originally Posted by thaopanda
I've been staring at this problem for quite some time...

Let z belong in C such that |z| = 1. Compute |1+z|^2 + |1-z|^2

If it was like |z + w|^2 where w was also a complex number, then that would be a lot like multiplying vectors using the dot product, correct? So it is much simpler where w is a real number? I'm not sure if I understand how to go about this...

- Nicole
Let $z = x + iy$ such that $|z| = 1$.

Therefore $x^2 + y^2 = 1$.

$1 + z = 1 + x + iy$ and $1 - z = 1 - x - iy$.

Therefore

$|1 + z|^2 = (1 + x)^2 + y^2$

$|1 - z|^2 = (1 - x)^2 + (-y)^2 = (1 - x)^2 + y^2$.

Therefore

$|1 + z|^2 + |1 - z|^2 = (1 + x)^2 + y^2 + (1 - x)^2 + y^2$

$= 1 + 2x + x^2 + 1 - 2x + x^2 + 2y^2$

$= 1 + 2x^2 + 2y^2$

$= 1 + 2(x^2 + y^2)$

$= 1 + 2$ since $x^2 + y^2 = 1$

$= 3$.

3. wow... so it was as straightforward as I thought it was.. I feel dumb now

Thank you very much!!
Nicole

4. Originally Posted by thaopanda
I've been staring at this problem for quite some time...

Let z belong in C such that |z| = 1. Compute |1+z|^2 + |1-z|^2

If it was like |z + w|^2 where w was also a complex number, then that would be a lot like multiplying vectors using the dot product, correct? So it is much simpler where w is a real number? I'm not sure if I understand how to go about this...

Remember that $|z|^2 = z\bar z$.
Therefore, $|1+z|^2+|1-z|^2 = (1+z)(1+\bar z) + (1-z)(1-\bar z)$.
Thus, $1+z+\bar z + z\bar z + 1 - z - \bar z + z\bar z = 1 + |z|^2 + |z|^2 = 1+1+1=3$.