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What is the angle between vectors z and -iz ?

Let z = x + iy = a_{1} + {ib}_{1} So - iz = - ix + y = y - ix = a_{2} + {ib}_{2} So angle between them is △ θ = (\tan^{- 1} (\frac{b_{1}}{a_{1}}) - \tan^{- 1} (\frac{b_{2}}{a_{2}})) = \tan^{- 1} (\frac{\frac{b_{1}}{a_{1}} - \frac{b_{2}}{a_{2}}}{1 + \frac{b_{2}}{a_{2}} \times \frac{b_{1}}{a_{1}}}) = \tan^{- 1} (\frac{\frac{y}{x} + \frac{x}{y}}{1 - \frac{y}{x} \times \frac{x}{y}}) = \tan^{- 1} (\frac{\frac{y}{x} + \frac{x}{y}}{0}) = \tan^{- 1} (\infty)' = \frac{π}{2}

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