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Work Shown:
(-i)^5 = (-1*i)^5
(-i)^5 = (-1)^5 * i^5
(-i)^5 = (-1)^5 * i^2*i^2*i
(-i)^5 = (-1)^5 * (-1)*(-1)*i
(-i)^5 = -1 * 1 * i
(-i)^5 = -i
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A shortcut to quickly computing i^5 is to note the remainder of 5/4 is 1, so this means that i^5 = i^1 = i. Another example is i^25 would equal the same thing since 25/4 has a remainder of 1 as well.
Answer:
[tex]\huge\boxed{(-i)^5=-i}[/tex]
Step-by-step explanation:
[tex]\sqrt{-1}=i\to i^2=\left(\sqrt{-1}\right)^2=-1\\\\a^n\cdot a^m=a^{n+m}\\\\(ab)^n=a^nb^n\\=========================\\\\\text{We have:}\\\\(-i)^5=(-i)^{2+2+1}=(-i)^2(-i)^2(-i)^1=(-1\cdot i)^2(-1\cdot i)^2(-i)\\\\=(-1)^2(i)^2(-1)^2(i)^2(-i)=(1)(-1)(1)(-1)(-i)=-i[/tex]
