Example Question - real part
Here are examples of questions we've helped users solve.
Analysis of Non-trivial Zeros of the Riemann Zeta Function
<p>The given statement in the image relates to the Riemann Hypothesis, which is a conjecture in complex analysis and number theory. The Riemann Hypothesis posits that all non-trivial zeros of the Riemann zeta function have a real part equal to \(\frac{1}{2}\). As of now, this hypothesis has neither been proven nor disproven and remains one of the most important unsolved problems in mathematics. Therefore, a solution or proof is not available.</p> <p>Note: It is important to point out that if by "solve the question" you are asking for a proof or disproof of the Riemann Hypothesis, this is currently impossible as it remains an open problem. Should you need any other type of assistance regarding the Riemann zeta function or related topics, please specify your query further.</p>
Simplified Evaluation of Complex Number Expression
The expression you're being asked to evaluate is: (1 + 3i)^8 + (1 - 3i)^8 This is an expression involving complex numbers. When raised to powers, complex numbers can sometimes simplify due to their periodic nature in the complex plane, but it is usually not feasible to compute high powers like 8 by hand without a significant amount of computation. In this case, there's a shortcut that can simplify the process using binomial expansion and the fact that i^2 = -1. To evaluate this expression, you would normally apply the binomial theorem, which states that: (a + b)^n = Σ[k=0 to n] (n choose k) * a^(n-k) * b^k Where "n choose k" is the binomial coefficient, calculated as: (n choose k) = n! / (k! * (n - k)!) However, for the powers of 8, this would involve a lot of terms and computation, namely: (1 + 3i)^8 = 1^8 + (8 choose 1)*1^7*(3i) + (8 choose 2)*1^6*(3i)^2 + ... + (3i)^8 (1 - 3i)^8 = 1^8 + (8 choose 1)*1^7*(-3i) + (8 choose 2)*1^6*(-3i)^2 + ... + (-3i)^8 You will notice that terms with odd powers of (3i) in (1 + 3i)^8 will cancel out with the corresponding terms in (1 - 3i)^8 due to the opposite signs. Only the even powers will remain, which are identical in both expressions because (-3i)^2n = (3i)^2n for all integers n, as both will be a real number. To simplify calculation, you can just compute the terms involving even powers of i from either (1 + 3i)^8 or (1 - 3i)^8 and then double the real part to get the final answer. However, for powers of 8 and without further tricks, this could be a tedious process and is typically done using a computer algebra system. For simplicity and practicality, let me assist you in solving this with such computational help: (1 + 3i)^8 + (1 - 3i)^8 simplifies to 2 * (1^8 + (8 choose 2)*1^6*(3i)^2 + (8 choose 4)*1^4*(3i)^4 + (8 choose 6)*1^2*(3i)^6 + (3i)^8). Each individual term can be computed, keeping in mind that i^2 = -1, i^4 = i^2 * i^2 = (-1)^2 = 1, i^6 = i^4 * i^2 = 1 * (-1) = -1, and i^8 = i^6 * i^2 = (-1) * (-1) = 1. Here's the simplification of the even-i-powered terms: (8 choose 2)(3i)^2 = 28 * 9 * (-1) = -252 (8 choose 4)(3i)^4 = 70 * 81 * 1 = 5670 (8 choose 6)(3i)^6 = 28 * 729 * (-1) = -20412 (3i)^8 = 6561 * 1 = 6561 Summing these and multiplying by 2 gives the real part: 2 * (1 - 252 + 5670 - 20412 + 6561) = 2 * (1 - 252 + 5670 - 20412 + 6561) = 2 * 51568 = 103136 So the final answer is: 103136
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