To prove the distributive property for complex numbers, we need to show that:

$t(w + z) = (tw) + (tz)$

Definitions:

• $t = r + is$
• $w = u + iv$
• $z = x + iy$

where $r, s, u, v, x, y \in \mathbb{R}$.

Step 1: Expand $w + z$

$w + z = (u + iv) + (x + iy) = (u + x) + i(v + y)$

Step 2: Multiply $t$ with $w + z$

$t(w + z) = (r + is)((u + x) + i(v + y))$

Using the distributive property of real numbers and $i^2 = -1$: $t(w + z) = r(u + x) + r i(v + y) + is(u + x) + i^2 s(v + y)$ $t(w + z) = r(u + x) + ir(v + y) + is(u + x) - s(v + y)$ $t(w + z) = \big(r(u + x) - s(v + y)\big) + i\big(r(v + y) + s(u + x)\big)$

Step 3: Expand $tw$ and $tz$ separately

1. $tw = (r + is)(u + iv)$: $tw = ru + riv + isu + i^2sv$ $tw = ru - sv + i(rv + su)$

2. $tz = (r + is)(x + iy)$: $tz = rx + riy + isx + i^2sy$ $tz = rx - sy + i(ry + sx)$

Step 4: Add $tw$ and $tz$

$tw + tz = \big((ru - sv) + i(rv + su)\big) + \big((rx - sy) + i(ry + sx)\big)$ $tw + tz = \big(ru - sv + rx - sy\big) + i\big(rv + su + ry + sx\big)$

Step 5: Compare $t(w + z)$ and $tw + tz$

1. From $t(w + z)$, we have: $t(w + z) = \big(r(u + x) - s(v + y)\big) + i\big(r(v + y) + s(u + x)\big)$

2. From $tw + tz$, we have: $tw + tz = \big(ru + rx - sv - sy\big) + i\big(rv + ry + su + sx\big)$

Both expressions are identical. Thus: $t(w + z) = tw + tz$

Conclusion:

The multiplication of complex numbers is distributive over addition.

Would you like a more detailed breakdown of any step or additional examples? Here are some related questions:

1. How do you prove associativity of complex number multiplication?
2. Why is $i^2 = -1$ crucial in these calculations?
3. Can you illustrate this property geometrically in the complex plane?
4. How does this distributive property relate to vector addition?
5. What happens if $t, w, z$ are purely imaginary numbers?

Tip: Always expand terms systematically to avoid missing negative signs, especially when working with $i^2 = -1$.