Question

If ( x + iy )( 3 – 4i ) = 5 + 12i, then find the value $\sqrt{{{x}^{2}}+{{y}^{2}}}$.

Answer 1

Hint: First we will multiply the two complex numbers and write it in the form of a + ib and then we will compare it with the right hand side of the equation and after that we find the value of x and y after that we will put it in the given equation to find it’s value.

Complete step-by-step answer:
So, let’s start by multiplying the two complex number,
$\begin{aligned} & \left( x+iy \right)\left( 3-4i \right) \\\ & =\left( 3x-4xi+3yi+4y \right) \\\ & =\left( 3x+4y \right)+\left( 3y-4x \right)i \\\ \end{aligned}$
Now we have converted it in the form of a + ib,
Now we will compare it with 5 + 12i.
Therefore, after comparing we get,
$\begin{aligned} & \left( 3x+4y \right)=5\text{ }..............\text{ (1)} \\\ & \left( 3y-4x \right)=12\text{ }..............\text{ (2)} \\\ \end{aligned}$
Now we will solve this two equations to find out the value of x and y:
Multiplying equation (1) by 4 and multiplying equation (2) by 3 and then adding them we get,
$\begin{aligned} & \left( 12x+16y \right)+\left( 9y-12x \right)=20+36 \\\ & 25y=56 \\\ & y=\dfrac{56}{25} \\\ \end{aligned}$
Now putting this in equation (2) we get,
$\begin{aligned} & 4x=3y-12 \\\ & 4x=3\left( \dfrac{56}{25} \right)-12 \\\ & x=\dfrac{42}{25}-3 \\\ & x=\dfrac{33}{25} \\\ \end{aligned}$
Now we have calculated both the values of x and y,
Now we will substitute these values in $\sqrt{{{x}^{2}}+{{y}^{2}}}$.
Hence, it’s value is:
$\begin{aligned} & =\sqrt{{{\left( \dfrac{56}{25} \right)}^{2}}+{{\left( \dfrac{33}{25} \right)}^{2}}} \\\ & =\dfrac{\sqrt{{{56}^{2}}+{{33}^{2}}}}{25} \\\ & =\dfrac{65}{25} \\\ & =\dfrac{13}{5} \\\ \end{aligned}$
Hence, the answer to this question is $\dfrac{13}{5}$.

Note: To solve this question one should be able to write any combination of complex numbers in the form of a + ib. While comparing one should be careful to choose which is the shortest way to solve those two equations to get the values of x and y.