Question

If \[\left( {x{\text{ }} + {\text{ }}iy} \right)\left( {2{\text{ }} + {\text{ }}cis\theta } \right){\text{ }} = {\text{ }}3\] then \[{x^2} + {y^2} - 4x{\text{ }} + {\text{ }}3\]= ?

Answer 1

Hint: While approaching these types of questions first go with separation of the values and then squaring. Before moving into the problem further first we will have to know what complex number and cisθ means. Complex number contains real and imaginary parts. In the equation x+iy, x is the real part and y is the imaginary part. Cisθ is the short form for the complex equation cosθ + isinθ

Complete step by step solution:
Given Equation \[\left( {x{\text{ }} + {\text{ }}iy} \right)\left( {2{\text{ }} + {\text{ }}cis\theta } \right){\text{ }} = {\text{ }}3\]
from this we have to calculate \[{x^2} + {y^2} - 4x{\text{ }} + {\text{ }}3\]
First we will be substituting the value \[cis\theta {\text{ }} = {\text{ }}cos\theta {\text{ }} + {\text{ }}isin\theta \] in the equation \[\left( {x{\text{ }} + {\text{ }}iy} \right)\left( {2{\text{ }} + {\text{ }}cis\theta } \right){\text{ }} = {\text{ }}3\]
i.e., \[\left( {x{\text{ }} + {\text{ }}iy} \right)\left( {2{\text{ }} + {\text{ }}cos\theta {\text{ }} + {\text{ }}isin\theta } \right){\text{ }} = {\text{ }}3\];
Simplifying the equation
\[\Rightarrow 2x{\text{ }} + {\text{ }}xcos\theta {\text{ }} + {\text{ }}i\left( {xsin\theta } \right){\text{ }} + {\text{ }}i\left( {2y} \right){\text{ }} + i\left( {ycos\theta } \right){\text{ }} + {\text{ }}{i^2}\left( {ysin\theta } \right){\text{ }} = {\text{ }}3\];
\[\Rightarrow 2x{\text{ }} + {\text{ }}xcos\theta {\text{ }} - ysin\theta {\text{ }} + {\text{ }}i\left( {xsin\theta {\text{ }} + 2y{\text{ }} + ycos\theta } \right) = 3\]
Since $i = \sqrt - 1, so i2= -1$
Now we have to equate real and imaginary parts on both the sides.
\[\Rightarrow 2x{\text{ }} + {\text{ }}xcos\theta {\text{ }} - ysin\theta {\text{ }} = 3\]; -------- say eq(1)
\[\Rightarrow xsin\theta {\text{ }} + {\text{ }}2y{\text{ }} + {\text{ }}ycos\theta {\text{ }} = {\text{ }}0\]; ------ say eq(2)

Rewriting the equations as the trigonometric and numerical values on either sides.
\[\Rightarrow xcos\theta {\text{ }} - ysin\theta {\text{ }} = 3{\text{ }} - 2x\]----- (1)
\[\Rightarrow xsin\theta {\text{ }} + {\text{ }}ycos\theta {\text{ }} = {\text{ }} - 2y\]----- (2)
Squaring on both sides of the equations
\[\Rightarrow {\left( {xcos\theta {\text{ }}-{\text{ }}ysin\theta } \right)^2} = {\text{ }}{\left( {3 - 2x} \right)^2}\]
\[\Rightarrow {x^2}co{s^2}\theta {\text{ }} + {y^2}si{n^2}\theta {\text{ }} - 2xysin\theta cos\theta {\text{ }} = 9{\text{ }} + {\text{ }}4{x^2} - 12x\]----- (1)
\[\
\Rightarrow {\left( {xsin\theta {\text{ }} + ycos\theta } \right)^2} = {\text{ }}{\left( { - 2y} \right)^2} \\
    \\
\ \]
\[\Rightarrow {x^2}si{n^2}\theta {\text{ }} + {y^2}co{s^2}\theta {\text{ }} + {\text{ }}2xysin\theta cos\theta {\text{ }} = {\text{ }}4{y^2}\]------ (2)
adding eq(1) and eq(2)
\[\Rightarrow {x^2}\left( {si{n^2}\theta {\text{ }} + {\text{ }}co{s^2}\theta } \right){\text{ }} + {\text{ }}{y^2}\left( {{\text{ }}si{n^2}\theta {\text{ }} + {\text{ }}co{s^2}\theta } \right){\text{ }} + {\text{ }}2xysin\theta cos\theta {\text{ }} - 2xysin\theta cos\theta = {\text{ }}9{\text{ }} + {\text{ }}4{x^2} - 12x{\text{ }} + {\text{ }}4{y^2}\]
we know that\[si{n^2}\theta {\text{ }} + {\text{ }}co{s^2}\theta {\text{ }} = {\text{ }}1\];
\[\
\Rightarrow {x^2} + {\text{ }}{y^2} = 9{\text{ }} + {\text{ }}4{x^2} - 12x{\text{ }} + {\text{ }}4{y^2} \\
\Rightarrow \;3{x^2} + {\text{ }}3{y^2} - 12x + 9 = 0 \\
\Rightarrow 3\left( {{x^2} + {\text{ }}{y^2}-{\text{ }}4x{\text{ }} + 3} \right) = 0 \\
\Rightarrow \;{x^2} + {\text{ }}{y^2}-{\text{ }}4x{\text{ }} + 3{\text{ }} = {\text{ }}0 \\
\ \]

Therefore \[{x^2} + {\text{ }}{y^2}-{\text{ }}4x{\text{ }} + 3{\text{ }} = {\text{ }}0.\]

Note: In these questions the whole idea depends on separating the imaginary and real parts and the further calculations. Make sure that you know the trigonometric identities and the operations involved in it. Many of the students go for the substituting in the question so avoid that an focus on how we can get the solution from squaring and adding.