2.1. BbIIIOJIHUTe LICTICTBHe: (5+i)/(2+3i) 2.2. BblyHCJIHTe: (1-i)^3 23. Bbluncjiute: (1+i)^3 2.4 . Bbruncjurre: (1+i)^4 2.5 Bbruncourre: i^20+i^12-i^10

1. $\frac {5+i}{2+3i}$<br />To simplify this expression, we can multiply both the numerator and denominator by the conjugate of the denominator, which is $2-3i$.<br />$\frac {5+i}{2+3i} \cdot \frac {2-3i}{2-3i} = \frac {(5+i)(2-3i)}{(2+3i)(2-3i)} = \frac {10-15i+2i-3i^2}{4-9i^2} = \frac {10-13i+3}{4+9} = \frac {13-13i}{13} = 1-i$<br /><br />2. $(1-i)^{3}$<br />To simplify this expression, we can expand it using the binomial theorem or by multiplying the expression by itself.<br />$(1-i)^{3} = (1-i)(1-i)(1-i) = (1-i)(1-2i+i^2) = (1-i)(1-2i-1) = (1-i)(-2i) = -2i+2i^2 = -2i-2 = -2-2i$<br /><br />3. $(1+i)^{3}$<br />To simplify this expression, we can expand it using the binomial theorem or by multiplying the expression by itself.<br />$(1+i)^{3} = (1+i)(1+i)(1+i) = (1+i)(1+2i+i^2) = (1+i)(1+2i-1) = (1+i)(2i) = 2i+2i^2 = 2i-2 = -2+2i$<br /><br />4. $(1+i)^{4}$<br />To simplify this expression, we can expand it using the binomial theorem or by multiplying the expression by itself.<br />$(1+i)^{4} = (1+i)(1+i)(1+i)(1+i) = (1+i)(1+2i+i^2)(1+i) = (1+i)(1+2i-1)(1+i) = (1+i)(2i)(1+i) = (1+i)(2i+2i^2) = (1+i)(2i-2) = (1+i)(-2+2i) = -2+2i-2i+2i^2 = -2+2i-2i-2 = -4$<br /><br />5. $i^{20}+i^{12}-i^{10}$<br />To simplify this expression, we can use the fact that $i^4 = 1$.<br />$i^{20}+i^{12}-i^{10} = (i^4)^5 + (i^4)^3 - (i^4)^2 \cdot i^2 = 1^5 + 1^3 - 1^2 \cdot i^2 = 1 + 1 - 1 \cdot (-1) = 1 + 1 + 1 = 3$