Question

If $\left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right)=A+iB$, then show that $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)=\left( {{A}^{2}}+{{B}^{2}} \right)$.

Answer 1

Hint: If we carefully notice the equation $\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)=\left( {{A}^{2}}+{{B}^{2}} \right)$, we will notice that it contains a term $\left( {{A}^{2}}+{{B}^{2}} \right)$ on the right side of the equality. Also, this term is the mod of the complex number $A+iB$ which is present in the equation $\left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right)=A+iB$.

Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question. In complex numbers, for a complex number $x+iy$, the mod $\left| x+iy \right|$ is given by the formula,
$\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}................\left( 1 \right)$
Also, there is a property of the mod function in the complex number. The mod function can be distributed over the multiplication of the two complex numbers. For example if we have two complex number $x+iy$ and $x'+iy'$ multiplied to each other, we can write,
$\left| \left( x+iy \right)\left( x'+iy' \right) \right|=\left| x+iy \right|\left| x'+iy' \right|.................\left( 2 \right)$
 In this question, we are given $\left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right)=A+iB$. If we take mod on both the sides of this equation, we get,
$\left| \left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right) \right|=\left| A+iB \right|$
From property $\left( 2 \right)$, we can distribute the mod function to each multiplied term. Using property $\left( 2 \right)$ in the above equation, we get,
$\left| \left( a+ib \right) \right|\left| \left( c+id \right) \right|\left| \left( e+if \right) \right|\left| \left( g+ih \right) \right|=\left| A+iB \right|$
Using formula $\left( 1 \right)$, we can simplify the above term by calculating the mod of each term.
$\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)}\sqrt{\left( {{c}^{2}}+{{d}^{2}} \right)}\sqrt{\left( {{e}^{2}}+{{f}^{2}} \right)}\sqrt{\left( {{g}^{2}}+{{h}^{2}} \right)}=\sqrt{\left( {{A}^{2}}+{{B}^{2}} \right)}$
We can also write the above equation as,
$\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)}=\sqrt{\left( {{A}^{2}}+{{B}^{2}} \right)}$
Squaring both the sides of the above equation, we get,
$\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)=\left( {{A}^{2}}+{{B}^{2}} \right)$
We have got the expression that was asked to prove in the question.
Hence proved.

Note: This question can also be solved by simply multiplying all the complex numbers in the equation $\left( a+ib \right)\left( c+id \right)\left( e+if \right)\left( g+ih \right)=A+iB$ and then applying the mod function on both the sides. But then, we have to factorize the final obtained expression which will take a lot of time. So, it is suggested to first apply mod function and then distribute the mod instead of multiplying the complex numbers and then applying mod function.