A Curious Rational Equation | Problem 371

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[@seanfraser3125]

2+i = i(-2i + 1) = i(1-2i) So (2+i/1-2i) = i(1-2i)/(1-2i) = i We thus have i^n = 1, giving the solution n=4k where k is any integer.

[@scottleung9587]

Got it!

[@mcwulf25]

Multiply top and bottom by (1+2i) and you get 5i/5 inside the brackets, or just i. So i^4k = 1.

[@lawrencejelsma8118]

Wow! 😮 Let me tell you about the story in my life when the eye doctor said I was now going to be 4 "i s" ...> ✊🤓👍 (i)^4 makes me "one of a kind!!!" 😂🤣

[@sdspivey]

I answered the problem, where's my million dollars?

[@key_board_x]

[(2 + i)/(1 - 2i)]^(n) = 1 [(2 + i).(1 + 2i)/(1 - 2i).(1 + 2i)]^(n) = 1 [(2 + 4i + i + 2i²)/(1 - 4i²)]^(n) = 1 [5i/(1 + 4)]^(n) = 1 [5i/5]^(n) = 1 i^(n) = 1 i^(n) = (- 1) * (- 1) → we know that: i² = - 1 i^(n) = i² * i² i^(n) = i⁴ n = 4 → but n = 8, n = 8, n = 12 and so on…

[@walterwen2975]

A Curious Rational Equation: [(2 + i)/(1 - 2i)]^n = 1; n =? (2 + i)/(1 - 2i) = [(2 + i)(1 + 2i)]/[(1 - 2i)(1 + 2i)] = (2 + 2i² + 5i)/(1 - 4i²) = (2 - 2 + 5i)/(1 + 4) = (5i)/5 = i, i^n = 1 = i⁴; n = 4 Answer check: [(2 + i)/(1 - 2i)]^n = 1; Confirmed as shown Final answer: n = 4