Complex Conjugates: I'm Not Sure Why V^2?

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SUMMARY

The discussion clarifies the relationship between complex numbers and their modulus, specifically addressing the expression VV* = |V|^2. The confusion arises from the incorrect assumption that V_r^2 + V_i^2 can be simplified to V^2 without considering the additional term from squaring a complex number. The correct expansion of V^2 = (a + ib)(a + ib) results in a^2 + 2iab - b^2, highlighting the importance of recognizing the imaginary component. Ultimately, the modulus of a complex number is accurately represented as sqrt(a^2 + b^2), and squaring this modulus yields the correct relationship.

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Statement:
[tex]VV* = (V_r + jV_i)(V_r - jV_i) = V_{r}^{2} + V_{i}^{2} = |V|^{2}[/tex]Question:
I am not sure why the second equality isn't written as, [tex]V_{r}^{2} + V_{i}^{2} = V^{2}?[/tex]
 
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Well no. Let's try V = a + ib and square it to see what happens.

V^2 = VV = (a + ib)(a + ib) = a^2 + aib + iba + (ib)^2 = a^2 + 2iab - b^2

Notice the extra 2iab term.

The modulus of a complex number is the length of the "vector" in the 2-d complex plane. Such vectors have x-component equal to the real part and y-component equal to the imaginary part. So, according to the Pythagorean theorem, they have modulus sqrt(a^2 + b^2). Squaring this gives the desired a^2 + b^2.
 

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