Half angle identity

[Albert Chan]

(04-17-2022 03:29 PM)Albert Chan Wrote:  \(\sqrt{2\;(x ± \sqrt{x^2-y^2})} = \sqrt{x+y}\;± \sqrt{x-y} \)

We can use the identity to build formula for complex square roots
Let Z = X+Y*i. For simplify assume Z on the unit circle.

\(\displaystyle \sqrt{2\;(X ± 1)} = \sqrt{Z}\;± \sqrt{\bar{Z}} \)

\(\displaystyle \sqrt{2Z}
= \sqrt{1+X} + i \; sgn(Y)\; \sqrt{1-X}
= \frac{(1+X)\;+\;i\;Y}{\sqrt{1+X}}
\)

\(\displaystyle \;\,\sqrt{Z} = \frac{Z+1}{|Z+1|}\)      // if |Z| = 1

Let Z = cis(θ). Matching parts of √Z = cis(θ/2), above gives half-angle formulas:
Assume θ = arg(z), with range ± pi

cos(θ/2) = (1+cos(θ)) / √(2+2 cos(θ)) = √((1+cos(θ))/2) ≥ 0
sin(θ/2) =     sin(θ)     / √(2+2 cos(θ)) = √((1−cos(θ))/2) * sgn(θ)