If \(\tan[\log(x+iy)]=a+ib\), prove that \(\tan[\log(x^2+y^2)]=\frac{2a}{1-a^2-b^2}\) when \(a^2+b^2\ne1\).

Let \(\log(x+iy)=\alpha+i\beta\). Then \(\alpha=\log\sqrt{x^2+y^2}=\frac12\log(x^2+y^2)\), so \(\log(x^2+y^2)=2\alpha\).

Given \(\tan(\alpha+i\beta)=a+ib\); taking conjugates, \(\tan(\alpha-i\beta)=a-ib\).

\(\tan2\alpha=\tan[(\alpha+i\beta)+(\alpha-i\beta)]=\frac{\tan(\alpha+i\beta)+\tan(\alpha-i\beta)}{1-\tan(\alpha+i\beta)\tan(\alpha-i\beta)}=\frac{(a+ib)+(a-ib)}{1-(a+ib)(a-ib)}=\frac{2a}{1-a^2-b^2}\)

\(\therefore\tan[\log(x^2+y^2)]=\frac{2a}{1-a^2-b^2}\), valid when \(a^2+b^2\ne1\) (proved).