Let z = \(\frac{1+\sqrt3}2=\frac12+\frac{\sqrt3}2i\)
\(\therefore\) a = \(\frac12\), b = \(\frac{\sqrt3}2\), a, b > 0
\(\therefore\) |z| = r = \(\sqrt{a^2+b^2}\)
Here, \((\frac12,\frac{\sqrt3}2)\) lies in 1st quadrant.
\(\therefore\) θ = amp(z) = tan-1(\(\frac{b}a\))