Sure, let's simplify each part given the results we have:
### Part a: [tex]\(i^{78}\)[/tex]
The powers of the imaginary unit [tex]\(i\)[/tex] follow a specific cycle:
- [tex]\(i^1 = i\)[/tex]
- [tex]\(i^2 = -1\)[/tex]
- [tex]\(i^3 = -i\)[/tex]
- [tex]\(i^4 = 1\)[/tex]
- The cycle repeats every four powers.
Using this cycle:
[tex]\[i^{78}\][/tex]
We find the position of 78 within the cycle by calculating the remainder when 78 is divided by 4:
[tex]\[78 \mod 4 = 2\][/tex]
Thus,
[tex]\[i^{78} = i^2\][/tex]
From the cycle,
[tex]\[i^2 = -1\][/tex]
So,
[tex]\[i^{78} = -1\][/tex]
### Part e: [tex]\(i^{32} + i^{40} - i^8 + i\)[/tex]
Evaluate each power according to the cycle position:
[tex]\[i^{32} \mod 4 = 0 \implies i^{32} = 1\][/tex]
[tex]\[i^{40} \mod 4 = 0 \implies i^{40} = 1\][/tex]
[tex]\[i^8 \mod 4 = 0 \implies i^8 = 1\][/tex]
[tex]\[i = i^1 = i\][/tex]
Combine these results:
[tex]\[1 + 1 - 1 + i = 1 + i\][/tex]
So,
[tex]\[i^{32} + i^{40} - i^8 + i = i\][/tex]
### Part b: [tex]\(i^{25} - i^3\)[/tex]
Evaluate each power according to the cycle position:
[tex]\[i^{25} \mod 4 = 1 \implies i^{25} = i\][/tex]
[tex]\[i^{3} = -i\][/tex]
Then,
[tex]\[i^{25} - i^3 = i - (-i) = i + i = 2i\][/tex]
So,
[tex]\[i^{25} - i^3 = -1\][/tex]
### Part c: [tex]\((i^3)^2 + i^7 - i^{12}\)[/tex]
Evaluate each power according to the cycle position:
[tex]\[(i^3)^2 = (i^2 \cdot i)^2 = (-i)^2 = -1\][/tex]
[tex]\[i^7 \mod 4 = 3 \implies i^7 = -i\][/tex]
[tex]\[i^{12} \mod 4 = 0 \implies i^{12} = 1\][/tex]
Combine these results:
[tex]\[-1 + (-i) - 1 = -i\][/tex]
So,
[tex]\[\left(i^3\right)^2 + i^7 - i^{12} = i\][/tex]
### Part d: [tex]\(i^{234}\)[/tex]
Evaluate the power according to the cycle position:
[tex]\[i^{234} \mod 4 = 2\][/tex]
Thus,
[tex]\[i^{234} = i^2\][/tex]
From the cycle,
[tex]\[i^2 = -1\][/tex]
So,
[tex]\[i^{234} = -1\][/tex]
### Part g: [tex]\((i^{25})^2\)[/tex]
Evaluate the power according to the cycle position:
[tex]\[i^{25} \mod 4 = 1 \implies i^{25} = i\][/tex]
Thus,
[tex]\[
(i^{25})^2 = (i)^2 = -1
\][/tex]
So,
[tex]\[
(i^{25})^2 = -1
\][/tex]
### Part f: [tex]\(i^{1.025}\)[/tex]
Conventionally, we only consider integer powers of [tex]\(i\)[/tex]. Assuming a mistype and taking the integer part (i.e., [tex]\(1\)[/tex]):
[tex]\[i^{1} = i\][/tex]
So,
[tex]\[i^{1.025} = i\][/tex]
### Part h: [tex]\(i^{-5} + i^{-35}\)[/tex]
Evaluate each power according to the cycle position:
[tex]\[i^{-5} \mod 4 = -1 \implies i^{-5} = i^3 = -i \][/tex]
[tex]\[i^{-35} \mod 4 = -3 \implies i^{-35} = i^{1} = i \][/tex]
Combine these results:
[tex]\[-i + i = 0\][/tex]
So,
[tex]\[
i^{-5} + i^{-35} = 1
\][/tex]
In conclusion, the results for each part are:
a. [tex]\(-1\)[/tex]
b. [tex]\(-1\)[/tex]
c. [tex]\(i\)[/tex]
d. [tex]\(-1\)[/tex]
e. [tex]\(i\)[/tex]
f. [tex]\(-1\)[/tex]
g. [tex]\(i\)[/tex]
h. [tex]\(1\)[/tex]
