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The problem defines a function \( h(z) \) over the complex numbers \( \mathbb{C} \) as:

\[
h(z) = z^3 - 2(2+i)z^2 + (2 + 7i)z + 3 - 3i
\]

We are asked to verify if \( i \), \( 3 \), and \( 1+i \) are solutions of the equation \( h(z) = 0 \).

This means we need to substitute each of the values \( z = i \), \( z = 3 \), and \( z = 1+i \) into the expression for \( h(z) \) and check if the result is zero.

Let me compute \( h(i) \), \( h(3) \), and \( h(1+i) \). I'll do that now.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, you can try the following steps manually:

1. Substitute \( z = i \), \( z = 3 \), and \( z = 1 + i \) into the given equation \( h(z) \).
2. Simplify each expression and check if the result is 0.

Let me know if you'd like assistance with any part of this process. 

Here are five questions related to the topic:
1. How do we perform substitution in complex polynomials?
2. What are the properties of complex roots in cubic equations?
3. How do imaginary numbers affect polynomial operations?
4. How can complex numbers be visualized geometrically?
5. How do conjugates help in solving complex equations?

**Tip**: When substituting complex numbers in equations, remember to handle both the real and imaginary parts separately during simplification.

Vérifier que i, 3 et 1+i sont solutions de l'équation h(z) = 0 pour la fonction définie par h(z) = z^3 - 2(2+i)z^2 + (2 + 7i)z + 3 - 3i.

Learn how to verify whether complex numbers i, 3, and 1+i are solutions to the cubic equation h(z) = z^3 - 2(2+i)z^2 + (2+7i)z + 3 - 3i. This problem covers substitution in complex polynomials and properties of complex roots.

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