The problem asks to simplify several expressions involving complex numbers. The expressions are: (a) $(2+i)(3-4i)$ (b) $(3+4i)(3-4i)$ (c) $(2-i)^2$ (d) $(1+i)^3$ (e) $(x+yi)(x-yi)$ (f) $i(1+i)(2+i)$ (g) $\frac{(1+i\sqrt{3})^2}{2}$

1. Problem Description

The problem asks to simplify several expressions involving complex numbers. The expressions are:

(a) $(2+i)(3-4i)$

(b) $(3+4i)(3-4i)$

(c) $(2-i)^2$

(d) $(1+i)^3$

(e) $(x+yi)(x-yi)$

(f) $i(1+i)(2+i)$

(g) $\frac{(1+i\sqrt{3})^2}{2}$

2. Solution Steps

(a) $(2+i)(3-4i) = 2(3) + 2(-4i) + i(3) + i(-4i) = 6 - 8i + 3i - 4i^2 = 6 - 5i - 4(-1) = 6 - 5i + 4 = 10 - 5i$

(b) $(3+4i)(3-4i) = 3(3) + 3(-4i) + 4i(3) + 4i(-4i) = 9 - 12i + 12i - 16i^2 = 9 - 16(-1) = 9 + 16 = 25$

Alternatively, using the difference of squares, $(3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25$

(c) $(2-i)^2 = (2-i)(2-i) = 2(2) + 2(-i) - i(2) + (-i)(-i) = 4 - 2i - 2i + i^2 = 4 - 4i - 1 = 3 - 4i$

Alternatively, $(2-i)^2 = 2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i$

(d) $(1+i)^3 = (1+i)(1+i)(1+i) = (1+i)(1+2i+i^2) = (1+i)(1+2i-1) = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$

Alternatively, $(1+i)^3 = (1+i)^2(1+i) = (1+2i+i^2)(1+i) = (1+2i-1)(1+i) = (2i)(1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$

(e) $(x+yi)(x-yi) = x^2 - (yi)^2 = x^2 - y^2i^2 = x^2 - y^2(-1) = x^2 + y^2$

(f) $i(1+i)(2+i) = i(2 + i + 2i + i^2) = i(2 + 3i - 1) = i(1 + 3i) = i + 3i^2 = i - 3 = -3 + i$

(g) $\frac{(1+i\sqrt{3})^2}{2} = \frac{1^2 + 2(1)(i\sqrt{3}) + (i\sqrt{3})^2}{2} = \frac{1 + 2i\sqrt{3} + i^2(3)}{2} = \frac{1 + 2i\sqrt{3} - 3}{2} = \frac{-2 + 2i\sqrt{3}}{2} = -1 + i\sqrt{3}$

3. Final Answer

(a) $10-5i$

(b) $25$

(c) $3-4i$

(d) $-2+2i$

(e) $x^2+y^2$

(f) $-3+i$

(g) $-1+i\sqrt{3}$