AMC 12 Mock Exam 4 Problem 10Problem 10 of 25What is the value of (1+i)2026(1−i)2022\dfrac{(1+i)^{2026}}{(1-i)^{2022}}(1−i)2022(1+i)2026, where i=−1i=\sqrt{-1}i=−1?Answer choicesA4441B8882C1616163D3232324E6464645Submit answerShow solutionSolutionSince 1+i1−i=(1+i)22=i\frac{1+i}{1-i}=\frac{(1+i)^2}{2}=i1−i1+i=2(1+i)2=i, (1+i)2026(1−i)2022=(1+i1−i)2022(1+i)4=i2022⋅(2i)2=(−1)(−4)=4.\frac{(1+i)^{2026}}{(1-i)^{2022}}=\left(\frac{1+i}{1-i}\right)^{2022}(1+i)^4=i^{2022}\cdot(2i)^2=(-1)(-4)=\boxed{4}.(1−i)2022(1+i)2026=(1−i1+i)2022(1+i)4=i2022⋅(2i)2=(−1)(−4)=4.Copy problemTagged: powers of complex numbers