**Theorem. **Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

In other words, if \(z = a+bi\) and \(w = c+di\), where \(a,b,c,d\) are real, and \(z = w \), then \(a = c\) and \(b = d\).

Let’s prove this fact.

**Proof. **It’s clear that if \(a = c\) and \(b = d\), then \(a + bi = c + di\).

If \(a + bi = c + di\), then \( a-c = (d-b)i \). Squaring both sides and collecting the terms gives us $$(a-c)^2 + (d-b)^2 = 0.$$ Now recall that the square of any real number is greater than or equal to zero. Thus, this is only possible if \(a-c = d-b = 0\), which means that \(a = c\) and \(b = d\).

## Problems

- Find \(x\) and \(y\), given that they are real numbers.
- \(4+5i = x+yi\)
- \(-2 -3i = x+(y-1)i\)
- \(2+i = (x-1) + (2y+3)i \)
- \(1+i =2x + yi + 4 \)
- \(2 + 3i = x + y + (x-y)i \)