What is the length of the vector (1,i) and why can it still be normalized?

[DocZaius]

I have a vector (1,i) and need to normalize it. I am being told that the answer is 1/(sqrt(2)) (1,i) but it seems clear to me that the vector's length is 0 and thus can't be normalized. What am i missing?

 

[I like Serena]

I have a vector (1,i) and need to normalize it. I am being told that the answer is 1/(sqrt(2)) (1,i) but it seems clear to me that the vector's length is 0 and thus can't be normalized. What am i missing?

Hi DocZaius!

The length of a vector containing complex numbers is defined slightly different.

From http://en.wikipedia.org/wiki/Norm_(mathematics):
On an n-dimensional complex space ℂⁿ the most common norm is
$$\|\boldsymbol{z}\| := \sqrt{|z_1|^2 + \cdots + |z_n|^2}= \sqrt{z_1 \bar z_1 + \cdots + z_n \bar z_n}.$$

 

[DocZaius]

Thanks!

 

[Fredrik]

The absolute value |z| of a complex number z=a+ib is defined by $|z|=\sqrt{a^2+b^2}$, or equivalently, by $|z|=\sqrt{z^*z}$, where z*=a-ib is the complex conjugate of z.

 

[blue_raver22]

The length of a vector is typically calculated using the Pythagorean theorem, which states that the length of a vector (a,b) can be found by taking the square root of the sum of the squares of its components, or √(a^2 + b^2). In this case, the vector (1,i) has components of 1 and i, so its length would be √(1^2 + i^2) = √(1 + (-1)) = √2. Therefore, the length of the vector (1,i) is indeed √2, not 0. The normalization process involves dividing each component of the vector by its length, so the normalized vector would be (1/√2, i/√2). This is equivalent to the answer you were given, 1/(√2) (1,i). It is important to note that the length of a vector is not the same as its magnitude, which is the absolute value of its components. In this case, the magnitude of the vector (1,i) is 1, but its length is √2. Therefore, there is no contradiction in the given answer and the length of the vector (1,i) can indeed be normalized.