Question

If$(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i,$then
A. $p=x,q=y$
B. $p={{x}^{2}},q={{y}^{2}}$
C. $x=q,y=p$
D. None of these

Answer 1

Type of question is based on the complex number. So we will use the properties of complex number to simplify the given equation i.e. $(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i,$ which will give us relation between $x,y,p$ and $q$.

Complete step-by-step solution:
Moving ahead with the question in step wise manner;
As we know that in complex numbers we have two values one is real and one is imaginary, real values are simple values with which the iota sign $'i'$ is not used. And with whom it is used termed as imaginary value.
Now we have equation $(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i,$ let us simplify it to find the relation between variables. So let us first solve the LHS side, by opening the brackets, so we will get;

& (x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i \\\ & xp+xqi+ypi+yq{{i}^{2}}=({{x}^{2}}+{{y}^{2}})i \\\ \end{aligned}$$ As we know that $${{i}^{2}}=1$$, so put the value of $${{i}^{2}}$$ in above equation and solve it further; $$\begin{aligned} & xp+xqi+ypi-yq=({{x}^{2}}+{{y}^{2}})i \\\ & \left( xp-yq \right)+\left( xq+yp \right)i=({{x}^{2}}+{{y}^{2}})i \\\ \end{aligned}$$ As we also know that to compare the two complex numbers equate the real value of a complex number with the real value of another complex number, and the same can be done with an imaginary number. In other words, equation real and imaginary value separately with other complex numbers. For example $$a+ib=c+id$$ then we can say that ‘a’ is equal to ‘c’ and ‘b’ is equal to ‘d’. So applying the same property in our equation, we have real and imaginary values in the LHS side that are $$xp-yq$$ and $$\left( xq+yp \right)$$ respectively. While in the RHS side of equation we have only imaginary value and no real value, i.e. imaginary value is $$\left( {{x}^{2}}+{{y}^{2}} \right)$$ and as there is no real value so it would be zero. So by using the comparison property of complex number, we can compare in our equation as, $$xp-yq$$ is equal to zero, because there is no real value in RHS side of equation, so we will get; $$\begin{aligned} & xp-yq=0 \\\ & x=\dfrac{yq}{p} \\\ \end{aligned}$$ So from here we got the value of ‘x’ in terms of other variables. Now similarly we can equate the imaginary value of equation, i.e. $$\left( xq+yp \right)$$ is equal to $$\left( {{x}^{2}}+{{y}^{2}} \right)$$, so we will get; $$xq+yp={{x}^{2}}+{{y}^{2}}$$ Here put the value of ‘x’ we got by equation the real value of complex number, and by solving we will get; $$\begin{aligned} & \left( \dfrac{yq}{p} \right)q+yp={{\left( \dfrac{yq}{p} \right)}^{2}}+{{y}^{2}} \\\ & \dfrac{y{{q}^{2}}}{p}+yp=\dfrac{{{y}^{2}}{{q}^{2}}}{{{p}^{2}}}+{{y}^{2}} \\\ & \dfrac{y({{q}^{2}}+{{p}^{2}})}{p}=\dfrac{{{y}^{2}}({{q}^{2}}+{{p}^{2}})}{{{p}^{2}}} \\\ \end{aligned}$$ Since both side we have $${{q}^{2}}+{{p}^{2}}$$ which will always be greater than zero so we can cancel them; so we will get; $$\begin{aligned} & \dfrac{y}{p}=\dfrac{{{y}^{2}}}{{{p}^{2}}} \\\ & \dfrac{{{y}^{2}}}{{{p}^{2}}}-\dfrac{y}{p}=0 \\\ & \dfrac{y}{p}\left( \dfrac{y}{p}-1 \right)=0 \\\ \end{aligned}$$ So from here we got two relations i.e. $$\dfrac{y}{p}$$ is equal to zero or equal to 1, so by having equal to zero we will get ‘y’ equal to zero, and by having $$\dfrac{y}{p}$$ equal to 1 we will get ‘y’ equal to ‘p’. Now put the value of ‘y’ in relation $$x=\dfrac{yq}{p}$$, so we will get, ‘x’ equal to ‘q’. So we got $$y=p,x=q$$; **Hence answer is option ‘c’ i.e. $$y=p,x=q$$.** **Note:** The other relation we got by having $$\dfrac{y}{p}$$ equal to zero, from here we got ‘y’ equal to zero. And if we keep the value of ‘y’ in the equation $$x=\dfrac{yq}{p}$$ we will get ‘x’ also equal to zero. Which is not any option. Hence option ‘c’ is correct.