Question:(a) Let $\mathbf{n}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{ib}$be a complex number, where aand barereal (positive or negative) numbers. Show that the product $\mathbf{nn}\mathbf{*}$is always a positive real number. (b) Let localid="1663080869595" $\mathbf{m}\mathbf{=}\mathbf{c}\mathbf{+}\mathbf{id}$be another complexnumber. Show that $\left|\mathrm{nm}\left|=\left|n\right|\right|m\right|$ .

The complex number n is

$n=a+ib$

where aand bare real numbers.

The complex number m is

$m=c+id$

Suppose $\mathit{x}\mathbf{=}\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$is a complex number then its complex conjugate can be given as $\overline{\mathbf{x}}\mathbf{=}\mathbf{x}\mathbf{*}\mathbf{=}\mathbf{a}\mathbf{-}\mathbf{ib}$

The given complex number is

$n=a+ib$

Its complex conjugate can be given as

$n=a-ib$

Therefore,

$\begin{array}{l}nn*=(a+ib)\left(a-ib\right)\\ nn*=a^2+iab-iab-i^2{b}^2\\ nn*=a^2-i^2{b}^2\\ nn*=a^2+b^2\end{array}$

From above we can see $nn*=a^2+b^2$, since $a$ and $b$are real numbers so, $nn*$is always a real number and positive number.

Hence it is proven that $nn*$is always a positive real number.

The given complex numbers are

$$\begin{gathered} n=a+ib \\ m=c+id \end{gathered}$$

Multiplying both the complex numbers

$\begin{array}{c}nm=(a+ib)(c+id)\\ =ac+ibc+i\mathrm{da}+i^{2}bd\\ =ac+ibc+ida-bd\\ =ac-bd+i(bc+da)\end{array}$

Taking modulus both sides

$\begin{array}{rcl}\left|nm\right|& =& \sqrt{{\left(ac-bd\right)}^2+{\left(bc+da\right)}^2}\\ & =& \sqrt{a^2{c}^2-2abcd+b^2{d}^2+b^2{c}^2+2abcd+a^2{d}^2}\\ & =& \sqrt{a^2{c}^2+b^2{d}^2+b^2{c}^2+a^2{d}^2}\\ & & \end{array}$

Since $n=a+ib$,so, $\left|n\right|=\sqrt{a^2+b^2}$

Since $m=c+id$,so,$\left|m\right|=\sqrt{c^2+d^2}$

Therefore,

$\left|n\right|\left|m\right|=\left(\sqrt{a^2+b^2}\right)\left(\sqrt{c^2+d^2}\right)=\sqrt{a^2{c}^2+b^2{d}^2+b^2{c}^2+a^2{d}^2}$

From the above calculation we can conclude that, $\left|mm\left|=\left|n\right|\right|m\right|$

Hence it is proven that $\left|mm\left|=\left|n\right|\right|m\right|$.