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**e^(i*pi)** Correct, take the first derivative of $\displaystyle y=x^3$ and solve $\displaystyle f'(1)$

The normal is always perpendicular to the tangent and so the product of their gradients must be -1: $\displaystyle m_1m_2 = -1$ where $\displaystyle m_1 = f'(1)$ = the gradient at that point and $\displaystyle m_2 = \frac{-1}{m_1}$

In both cases use the equation of a straight line formula:

$\displaystyle y-y_1 = m(x-x_1)$ where $\displaystyle (x_1,y_1)$ is a coordinate. In your case this is (1,1)