Question

Find the value of the \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}\], If (\[x+iy=\sqrt{\dfrac{a+ib}{c+id}}\].
(a). \[\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}\]
(b). \[\dfrac{a+b}{c+d}\]
(c).\[\dfrac{{{c}^{2}}+{{d}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
(d). \[{{\left( \dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}\]

Answer 1

Hint: In the given expression, replace i, by (-i). Now multiply both the equations. Use basic identities like \[\left( {{a}^{2}}-{{b}^{2}} \right)\] and solve it. Finally square the LHS and RHS of the equation formed.

Complete step-by-step solution -

Given to us an expression of complex numbers, let us mark it as (1).
\[x+iy=\sqrt{\dfrac{a+ib}{c+id}}-(1)\]
Now let us replace i by (-i) in the above equation, we get
\[x-iy=\sqrt{\dfrac{a-ib}{c-id}}-(2)\]
Now let us multiply equation (1) and (2).
\[\left( x+iy \right)\left( x-iy \right)=\sqrt{\dfrac{a+ib}{c+id}}\times \sqrt{\dfrac{a-ib}{c-id}}\]
We know the formula, \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. Apply the same in the above expression.
\[{{x}^{2}}-{{\left( iy \right)}^{2}}=\sqrt{\dfrac{\left( a+ib \right)}{\left( c+id \right)}\times \dfrac{\left( a-ib \right)}{\left( c-id \right)}}\left\{ \because {{i}^{2}}=-1 \right\}\]
\[\begin{align}
  & {{x}^{2}}-{{i}^{2}}{{y}^{2}}=\sqrt{\dfrac{{{a}^{2}}-{{\left( ib \right)}^{2}}}{{{c}^{2}}-{{\left( id \right)}^{2}}}} \\
 & {{x}^{2}}-\left( -1 \right){{y}^{2}}=\sqrt{\dfrac{{{a}^{2}}-\left( -1 \right){{b}^{2}}}{{{c}^{2}}-\left( -1 \right){{d}^{2}}}} \\
\end{align}\]
Thus we get,
\[{{x}^{2}}+{{y}^{2}}=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}}-(3)\]
Now let us square both sides of equation (3).
\[\begin{align}
  & {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( \sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}} \right)}^{2}} \\
 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}} \\
\end{align}\]
Hence we got the value of \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}\].
\[\therefore \] Option (a) is the correct answer.

Note: You may also try to solve it by taking conjugate of \[\left( a+ib \right)\] then multiplying it on numerator and denominator. Then comparing the real and imaginary part of the complex method, you can also do it this way but it's lengthy and complicated. Thus we use the above method.