We have $\tan \big [ \log (x+ iy) \big ] = a+ ib$

$\therefore \big [ \log (x+ iy) \big ] = \tan ^{-1} (\alpha + ib)$

Let $\tan ^{-1} (\alpha + ib) = \alpha + i\beta \ \cdots \ \cdots (1)$

$\therefore \log (x+ iy) = \alpha + i\beta$

$\therefore \log \sqrt{(x^2 + y^2)} + i \tan ^{-1} \left ( \dfrac {y}{x} \right ) = \alpha + i \beta$

Equating real parts, we have

$\alpha = \log \sqrt{(x^2 + y^2)} $ i.e. $2 \alpha = \log (x^2 + y^2 ) \ \cdots \ \cdots (2)$

We now find the real part α of (1)

$\therefore \tan 2 \alpha = \tan \big [ (\alpha + i \beta) + \tan (\alpha - i \beta ) \big ]$

$\therefore \tan (2 \alpha) = \dfrac {\tan (\alpha + i \beta ) + \tan (\alpha - i\beta)}{1- \tan (\alpha + i\beta) \tan (\alpha - i \beta)} \ \cdots \ \cdots (3)$

Since $\tan^{-1} (\alpha + i b) = \alpha + i\beta$

$\therefore \tan (\alpha + i \beta ) = \alpha + i b \text { and } \tan (\alpha - i \beta ) = \alpha - i b$

Hence from (3) we get.

$\therefore \tan (2\alpha) = \dfrac {\alpha + ib + \alpha - ib}{1- (\alpha + ib) (\alpha - ib)} = \dfrac {2\alpha }{1-\alpha^2 - b^2}$

$\therefore 2\alpha =\tan^{-1} \left ( \dfrac {2\alpha}{1-\alpha^2 - b^2} \right ) \ \cdots \ \cdots (4)$

From (2) and (4)

$\tan \big [ \log (x^2 + y^2 \big ] = \dfrac {2\alpha }{1-\alpha^2 - b^2}$