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I have not dealt with complex numbers for a while now, but I was wondering if I multiplied the complex number $a+bi$ by its conjugate $a-bi$ to obtain

$$(a+bi)(a-bi) = a^2 + b^2 $$

where $a,b \in \mathbb{R}$, does $a^2 + b^2 \in \mathbb{R}$ or $\mathbb{C}$?

I am pretty sure that the above product is solely a real number since the imaginary number $i$ has been removed, but I want to make sure.

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4 Answers

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Yes, it is always in $\mathbb R$. In fact it is the square of the length of the vector pointing to the point in the complex plane.

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Think of this way: let $z_1 = r_1 \angle \theta_1$ and $z_2 = r_2 \angle \theta_2$, then $z_1 z_2 = r_1 r_2 \angle (\theta_1+\theta_2)$.

Now a picture saves the day!

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Yes it is a real number. But it is also a complex number as well.

$$\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$$

$a^2+b^2$ is a complex number where the imaginary part is $0$, i.e., Im$(a^2+b^2)=0$.

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$$|(a+bi)|=\sqrt{a^2+b^2}$$ $$|(a+bi)|=\sqrt{(a+bi)(a-bi)} = \sqrt{a^2 + b^2} $$

So yes it is real, for example pick a=4 & b=5:

$$|(4+5i)|=\sqrt{4^2+5^2}=\sqrt{(4+5i)(4-5i)}=\sqrt{41}$$

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