Find the principal argument of (1+i\(\sqrt{3}\))2

1.	Find the principal argument of (1+i\(\sqrt{3}\))2
Answer» As we know that, z = a+ib z = (1+i\(\sqrt{3}\))² = 1² + \((\sqrt{3i})^2\) + 2 x 1 x \(\sqrt{3i}\) = 1+i²+2i√3 = 1-3+2i√3 = -2+2i√3 a = -2 b = 2√3 tanα = \(\|\frac{b}{a}\|\) = \(\|\frac{2\sqrt{3}}{-2}\|\) = \|√3\| = α = \(\frac{\pi}{3}\) or 60° α<0,b>1 ∴z lies in second quadrant arg(z) = θ =π - a = π = - \(\frac{π}{3}\) = \(\frac{2π}{3}\)

Answer»

As we know that, z = a+ib

z = (1+i\(\sqrt{3}\))²

= 1² + \((\sqrt{3i})^2\) + 2 x 1 x \(\sqrt{3i}\)

= 1+i²+2i√3

= 1-3+2i√3

= -2+2i√3

a = -2 b = 2√3

tanα = \(|\frac{b}{a}|\)

= \(|\frac{2\sqrt{3}}{-2}|\)

= |√3|

= α = \(\frac{\pi}{3}\) or 60°

α<0,b>1

∴z lies in second quadrant

arg(z) = θ

=π - a

= π = - \(\frac{π}{3}\)

= \(\frac{2π}{3}\)

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