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

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SUMMARY

The length of the vector (1,i) is calculated using the norm for complex numbers, defined as ||z|| = √(|z_1|² + |z_2|²). For the vector (1,i), this results in a length of √(1² + 1²) = √2. Consequently, the normalized vector is (1/√2)(1,i), which is valid despite the initial misconception that its length is zero. The confusion arises from misunderstanding the definition of vector length in complex spaces.

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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?
 
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DocZaius said:
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! :smile:

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
$$\|\boldsymbol{z}\| := \sqrt{|z_1|^2 + \cdots + |z_n|^2}= \sqrt{z_1 \bar z_1 + \cdots + z_n \bar z_n}.$$
 
Thanks!
 
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.
 

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