Using Lagrange interpolation, find the unique polynomial $p(z)\in \mathbb{P}_3(\mathbb{C})$ such that
$p(1)=i, p(i)=1, p(−1)=1, p(−i)=−i.$ (The coefficients of this polynomial should be written in standard form $a+ib.$) I know how to solve this in other way ,but I am not sure about Lagrange interpolation
The process is exactly the same as in the real case and the Lagrange polynomials are:
$$\begin{align}L_0(x) &= \dfrac{(x+1)(x-i)(x+i)}{(1-i)(1+1)(1+i)}\\L_1(x)& = \dfrac{(x-1)(x-i)(x+i)}{(-1-1)(-1-i)(-1+i)}\\L_2(x) &= \dfrac{(x-1)(x+1)(x+i)}{(i-1)(i+1)(i+i)}\\L_3(x) &= \dfrac{(x-1)(x+1)(x-i)}{(-i-1)(-i+1)(-i-i)} \end{align}$$
The final result is given by
$$P_3(x) = (i)^3(x)+(1)^3L_1(x)+(1)^3L_2(x)+(-i)^3L_3(x)$$
This results in
$$P_3(x) = -(0.5\, -0.5 i) x^3+(0.\, +0.5 i) x^2+(0.5\, +0. i)$$