Answer: a) i-1; b) 4+i/13
Step-by-step explanation:
The complex number [tex]i[/tex] is defined as the number such that [tex]i^{2}=-1[/tex];
We use the propperty to notice that [tex]i^1=i \quad i^2 = -1 \quad i^3=-i \quad i^4=1 \quad i^5=1 \quad i^6=-1 \quad i^7=-i \quad i^8=1 \quad i^9=i \quad i^10= -1 etc...[/tex].
a) We notice that [tex]i^{21}=i \quad \text{ and \text} \quad i^{32}=1[/tex]. Hence, [tex]i^{21}-i^{32}=i-1[/tex].
b) We multiply the expression by [tex]1=\frac{3+2\cdot i}{3 + 2 \cdot i}[/tex]. Then we get that
[tex]\frac{2-i}{3-2 \cdot i}=\frac{2-i}{3-2\cdot i}\cdot\frac{3+2 \cdot i}{3 + 2\cdot i } = \frac{(2-i)\cdot(3+2 \cdot i)}{3^2+2^2}= \frac{6+4i-3i+2i^2}{13}=\frac{6+i-2}{13} = \frac{4+i}{13}[/tex]
