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Solution of |Z| - Z = 1 + 3i will be:1. 4 + 3i2. 3 - 4i3. 4 - 3i4. 3 + 4i |
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Answer» Correct Answer - Option 3 : 4 - 3i Concept: Let, Z1 = a1 + jb1 and Z2 = a2 + jb2 Z1 and Z2 are said to equal when, a1 = a2 and b1 = b2 Calculation: Let, Z = x+ iy Given that, |Z| - Z = 1 + 3i ⇒ |x + iy| - (x + iy) = 1 + 3i \(⇒\ \sqrt{x^2\ +\ y^2}\ -\ x\ -\ iy\ =\ 1\ +\ 3i\) Comparing real and imaginary parts of both sides, we will get \(⇒\ \sqrt{x^2\ +\ y^2}\ -\ x\ =\ 1\) -----(1) - i y = 3i ⇒ y = - 3 Therefore, from equation (1) \(\ \sqrt{x^2\ +\ (-3)^2}\ -\ x\ =\ 1\) \(⇒ \ \sqrt{x^2\ +\ 9}\ \ =\ 1 + x\) Taking square both side x2 + 9 = (1 + x)2 but we know that, (a + b)2 = a2 + 2ab + b2 ⇒ x2 + 9 = 1 + x2 + 2x ⇒ x = 4 Therefore, solution of given equation will be Z = x + iy = 4 - 3i Hence, option 3 is correct. |
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