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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 = b2. a = -b3. a = 2b4. a = b |
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Answer» 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. |
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