Question

A real and positive value of a and b will satisfy the equation \(\sqrt{2ab}(\frac{Z}{\bar{Z}})\ =\ a + ib\), Z = (b + ia) if:
1. 2a = b
2. a = -b
3. a = 2b
4. a = b

Answer 1

Correct Answer - Option 4 : a = b

Concept:

A complex number (Z):  Complex number is the combination of a real number and an imaginary number. It is given by

Z = x + iy, where 'x' and 'y' are the real and imaginary part of Z and i = √-1 

Conjugate of a complex number: When the i of a complex number is replaced with - i, we get the conjugate of that complex number.

\(\bar{Z}\ =\ x\ -\ iy\)

Re(Z) = x

\Img(Z) = y

|Z| = \(\sqrt{x^2\ +\ y^2}\)

Formula used:

1. \( |\frac{Z_1}{Z_2}| = \frac{|Z_1|}{|Z_2|}\)

2. \(Z\bar{Z}\ = |Z|^2\)

3. (a - b)² = a2 + b2 - 2ab

Calculation:

Given that,

\(\sqrt{2ab}(\frac{Z}{\bar{Z}})\ =\ a + ib\)     -----(1)

Z = (b + ia)     ----(2)

Therefore, a conjugate of Z

Z̅ = b - ia      ----(3)

Hence, from equation (1)

\(\sqrt{2ab}(\frac{b\ +\ ia}{b\ -\ ia})\ =\ a + ib\)

Taking modulus of both sides,

\(\sqrt{2ab}|(\frac{b\ +\ ia}{b\ -\ ia})|\ =\ |a + ib|\)

\(\sqrt{2ab}\frac{|b\ +\ ia|}{|b\ -\ ia|}|\ =\ |a + ib|\)    

\(\sqrt{2ab}\frac{\sqrt{b^2\ +\ a^2}}{\sqrt{b^2\ +\ a^2}​​​​}\ = \sqrt{a^2\ +\ b^2}\)   

Taking square of both side

a² + b² - 2ab = 0

⇒ (a - b)² = 0

⇒ a = b

Hence, option 4 is correct.