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What is the principal argument of \(\frac{1}{1 + i}\) where \(i = \sqrt{-1}?\)
Concept:Let z = x + iy be a complex number.QuadrantSign of x and yPrinciple value of ArgumentIx > 0, y > 0\(\rm \tan^{-1} \frac{y}{x}\)IIx < 0, y > 0? - \(\rm \tan^{-1} \left|\frac{y}{x}\right|\)IIIx < 0, y < 0-? + \(\rm \tan^{-1} \left|\frac{y}{x}\right|\)IVx > 0, y < 0\(-\rm \tan^{-1} \left|\frac{y}{x}\right|\) Calculation:Let, x + iy = \(\frac{1}{1 + i}\)? x + iy = \(\frac{1}{1 + i}\times \frac{1-i}{1-i}\)We know that, (a - b)(a + b) = a2 - b2? x + iy = \(\frac{1-i}{1 - i^2}\)? x + iy = \(\frac{1-i}{2} = \frac{1}{2}-\frac{i}{2}\) [? i = ?-1]Compare the real and imaginary part? x = 1/2 & y = -1/2So, x > 0 and y < 0Therefore, principal argument \(-\rm \tan^{-1} \left|\frac{y}{x}\right|\)? \(-\rm \tan^{-1} \left|\frac{-1/2}{1/2}\right|=-tan^{-1}1\)? - ?/4 ? The principal argument of \(\frac{1}{1 + i}\) is \(-\frac{\pi}{4}\).
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