It is impossible to eliminate both the \(x^5\) term and the \(x^6\) term at once, so we set \(b = 0 \rightarrow f(x) + b \cdot g(x) = f(x) = 2x^5 - 6x^4 - 4x^3 + 12x^2 + 7x + 5\). \(\text{deg}(f(x) + b \cdot g(x) = \boxed{5})\)
Honestly, you're lucky to get complete solutions at all. We're doing your work for free, and more demands aren't exactly sufficient incentives for proffering full solutions.
Because \((i, i^2, i^3, i^4) = (i, -1, -i, 1)\), we have \(i + i^2 + i^3 \cdots + i^{258} + i^{259} = 64(i - 1 - i + 1) + i^{257} + i^{258} + i^{259}\). So, the desired sum is \(i^{257} + i^{258} + i^{259} = i - 1 - i = \boxed{-1}\).
If \(z = a + bi\), then \(\overline{z} = a - bi\). So, \(z \overline{z} = (a + bi)(a-bi) = a^2 + b^2\). Since \((a, b) = (3, 5)\), we have \(z \overline{z} = 3^{2} + 5^{2} = 41.\)
You published the same question twice. Both times, you failed to provide a diagram. As such, it is impossible for any of us to help you with your problem.
Hint: the number of ways to distribute \(n\) indistinguishable balls into \(k\) distinguishable boxes is found with the binomial coefficient \(\binom{n + k - 1}{n}\).
https://web2.0calc.com/questions/sequence_123
