
If \[\omega \] is a complex number satisfying\[\mid \;\omega + \dfrac{1}{\omega }\;\mid = 2\], then maximum distance of \[\omega \] from origin is
A) \[2 + \sqrt 3 \]
B) \[1 + \sqrt 2 \]
C) \[1 + \sqrt 3 \]
D) None of these
Answer
162.6k+ views
Hint: in this question we have to find maximum distance of \[\omega \]from the origin. So here in this question simply find the modulus of \[\omega \]in order to find maximum distance. Solve quadratic equation formed by \[\omega \]by using Sri Dharacharya rule and take positive value for maximum distance.
Formula Used: General form of quadratic equation is given by
If \[a{x^2} + bx + c = 0\]
Root of this equation is given by Sri Dharacharya method
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where
b is coefficient of x
a is coefficient of \[{x^2}\]
c is constant
Complete step by step solution: Given: Equation in the form of complex number
Now we have complex number equation \[\mid \;\omega + \dfrac{1}{\omega }\;\mid = 2\]
\[\left| \omega \right| = \left| {\omega + \dfrac{1}{\omega } - \dfrac{1}{\omega }} \right| \le \left| {\omega + \dfrac{1}{\omega }} \right| + \left| {\dfrac{1}{\omega }} \right|\]
\[\left| \omega \right| = \left| {\omega + \dfrac{1}{\omega }} \right| + \left| {\dfrac{1}{\omega }} \right| = 2 + \dfrac{1}{{\left| \omega \right|}}\]
On simplification
\[\left| \omega \right| = 2 + \dfrac{1}{{\left| \omega \right|}}\]
\[{\left| \omega \right|^2} - 2\left| \omega \right| - 1 \le 0\]
Solve by using Sri Dharacharya method
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[\left| \omega \right| \le \dfrac{{2 \pm 2\sqrt 2 }}{2}\]
For maximum distance we take positive value
\[\left| \omega \right| = \dfrac{{2 + 2\sqrt 2 }}{2}\]
Now required distance is given as
\[\left| \omega \right| = 1 + \sqrt 2 \]
Option ‘B’ is correct
Note: Don’t try to put the value of \[\omega \] in combination of real and imaginary form. Just simplify the given equation. Here \[\omega \]form quadratic equation and solution quadratic equation is given by Sri Dharacharya method. Remember that for maximum distance take positive value.
Complex number is a number which is a union of real and imaginary number. So in complex number questions, we have to represent number as a combination of real and its imaginary part in most of the question. Imaginary part is known as iota. Square of iota is equal to negative one.
Formula Used: General form of quadratic equation is given by
If \[a{x^2} + bx + c = 0\]
Root of this equation is given by Sri Dharacharya method
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Where
b is coefficient of x
a is coefficient of \[{x^2}\]
c is constant
Complete step by step solution: Given: Equation in the form of complex number
Now we have complex number equation \[\mid \;\omega + \dfrac{1}{\omega }\;\mid = 2\]
\[\left| \omega \right| = \left| {\omega + \dfrac{1}{\omega } - \dfrac{1}{\omega }} \right| \le \left| {\omega + \dfrac{1}{\omega }} \right| + \left| {\dfrac{1}{\omega }} \right|\]
\[\left| \omega \right| = \left| {\omega + \dfrac{1}{\omega }} \right| + \left| {\dfrac{1}{\omega }} \right| = 2 + \dfrac{1}{{\left| \omega \right|}}\]
On simplification
\[\left| \omega \right| = 2 + \dfrac{1}{{\left| \omega \right|}}\]
\[{\left| \omega \right|^2} - 2\left| \omega \right| - 1 \le 0\]
Solve by using Sri Dharacharya method
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[\left| \omega \right| \le \dfrac{{2 \pm 2\sqrt 2 }}{2}\]
For maximum distance we take positive value
\[\left| \omega \right| = \dfrac{{2 + 2\sqrt 2 }}{2}\]
Now required distance is given as
\[\left| \omega \right| = 1 + \sqrt 2 \]
Option ‘B’ is correct
Note: Don’t try to put the value of \[\omega \] in combination of real and imaginary form. Just simplify the given equation. Here \[\omega \]form quadratic equation and solution quadratic equation is given by Sri Dharacharya method. Remember that for maximum distance take positive value.
Complex number is a number which is a union of real and imaginary number. So in complex number questions, we have to represent number as a combination of real and its imaginary part in most of the question. Imaginary part is known as iota. Square of iota is equal to negative one.
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