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

[jeff1evesque]

Statement:
$$VV* = (V_r + jV_i)(V_r - jV_i) = V_{r}^{2} + V_{i}^{2} = |V|^{2}$$Question:
I am not sure why the second equality isn't written as, $$V_{r}^{2} + V_{i}^{2} = V^{2}?$$

 

[AUMathTutor]

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.

 

[tomcue]

The second equality is written as V_{r}^{2} + V_{i}^{2} = |V|^{2} because V^{2} is not the same as |V|^{2}. V^{2} refers to the square of a single complex number, while |V|^{2} represents the magnitude squared of a complex number. In other words, |V|^{2} takes into account both the real and imaginary components of the complex number, while V^{2} only considers the magnitude. Therefore, it is important to use |V|^{2} in the context of complex conjugates, as it accurately represents the relationship between the real and imaginary components of the complex number.