Question

Compute the following quantities: a) i^21-i^32

b) 2-i/3-2i.

Please show work

Answer 1

Answer: a) i-1; b) 4+i/13

Explanation:

The complex number
`i` is defined as the number such that
`i^(2)=-1`;

We use the propperty to notice that
`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...`.

a) We notice that
`i^(21)=i \quad \text{ and \text} \quad i^(32)=1`. Hence,
`i^(21)-i^(32)=i-1`.

b) We multiply the expression by
`1=(3+2\cdot i)/(3 + 2 \cdot i)`. Then we get that

`(2-i)/(3-2 \cdot i)=(2-i)/(3-2\cdot i)\cdot(3+2 \cdot i)/(3 + 2\cdot i ) = ((2-i)\cdot(3+2 \cdot i))/(3^2+2^2)= (6+4i-3i+2i^2)/(13)=(6+i-2)/(13) = (4+i)/(13)`