Subjective Type

(a) Let $$n=a+i b$$ be a complex number, where $$a$$ and $$b$$ are real (positive or negative) numbers. Show that the product $$n \overline n$$ is always a positive real number. (b) Let $$m=c+i d$$ be another complex number. Show that $$|n m|=|n||m|$$

Solution

(a) The product $$n \overline n$$ can be rewritten as
$$\begin{aligned}n \overline n &=(a+i b)\overline {(a+i b)}=(a+i b)(a-i b) \\&=a^{2}+i b a-i a b+(i b)(-i b)=a^{2}+b^{2}\end{aligned}\\$$
which is always real since both $$a$$ and $$b$$ are real.
(b) Straightforward manipulation gives
$$\begin{aligned}|n m| &=|(a+i b)(c+i d)|=\left|a c+i a d+i b c+(-i)^{2} b d\right|=|(a c-b d)+i(a d+b c)| \\&=\sqrt{(a c-b d)^{2}+(a d+b c)^{2}}=\sqrt{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}\end{aligned}\\$$
However, since
$$\begin{aligned}|n||m| &=|a+i b||c+i d|=\sqrt{a^{2}+b^{2}} \sqrt{c^{2}+d^{2}} \\&=\sqrt{a^{2} c^{2}+b^{2} d^{2}+a^{2} d^{2}+b^{2} c^{2}}\end{aligned}\\$$
we conclude that $$|n m|=|n||m|$$

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