Let i be the imaginary unit. What complex number in standard form does the following simplify to: i^(425)+i^(14)+i^(-14)+i^(44) Basically you are being asked to find A and B suc…

Question

Let
`i` be the imaginary unit.

What complex number in standard form does the following simplify to:


`i^(425)+i^(14)+i^(-14)+i^(44)`

Basically you are being asked to find A and B such that the following equality holds for real values A and B:

`A+Bi=i^(425)+i^(14)+i^(-14)+i^(44)`

You must show all work.
Have fun.

Answer 1

`i - 1`

Explanation:

Complex Number System Rules

`√(-1) = i \\ -1 = {i}^(2) \\ -i = {i}^(3) \\ 1 = {i}^(4) [Every \: multiple \: of \: four]`

The Divisibilty Rule of 4 states that the last two digits of a number must be a multiple of four, so you will have this:

`{i}^(425) + {i}^(14) + {i}^(-14) + {i}^(44) \\ \\ i - 1 - 1 + 1 = i - 1`

I am joyous to assist you anytime.

Answer 2

`Hey~freckledspots!\\----------------------`

`We~will~solve~for~i^(425)!`

`Rule~of~exponent: a^(b + c) = a^ba^c\\Apply:~i^(425)~=~i^(424)i\\ \\Rule~of~exponent: a^(bc) = (a^(b))^c\\Apply: i^(424) = i(i^2)^(212) \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^(212) = -1^(212)i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^(212)i = 1^(212)i\\\\Simplify: 1^(212)i = 1i\\Multiply: 1i * 1 = i\\----------------------\\`

`Now~let's~solve~1^(14)!\\\\Rule~of~exponent: a^(b + c) = a^ba^c\\Apply: i^(14) = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^(-14)+i^(44)\\----------------------`

`Now~lets~solve~i^(-14)\\\\Rule~of~exponent: a^(-b) = (1)/(a^b) \\Apply: i^(-14) = (1)/(i^(14)) \\\\Rule~of~exponent: a^(bc) = (a^b)^c\\Apply: (1)/(i^(14)) = (1)/((i^2)^7)\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: (1)/((i^2)^7) = (1)/(-1^7) \\\\Simplify: (1)/(-1^7) = (1)/(-1) \\\\Rule~of~fractions: (a)/(-b) = -(a)/(b) \\Apply: (1)/(-1) = -(1)/(1) = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------`

`Now~let's~solve~i^(44)!\\\\Rule~of~exponent: a^(bc) = (a^b)^c\\Apply: i^(44) = (i^2)^(22)\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^(22) = -1^(22)\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^(22) = 1^(22)\\\\Simplify: 1^(22) = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------`

`Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------`

`Answer:\\\large\boxed{-1+i}\\----------------------`

`Hope~This~Helped!~Good~Luck!`