The correct simplifications of powers of i are:
a. i^27=−i
d. i^8=1
e. i^5=i
Step-by-step explanation:
We know that
[tex]i^2=-1\\i^3=-i\\i^4=1[/tex]
We can use the smaller powers of i to solve the larger powers.
a. i^27=−i
[tex]i^{27}\\={i^{24}}.i^3\\={i^{4*6}}.-i\\=(i^4)^6.-i\ \ \ \ As\ we\ know\ i^4=1\\=(1)^6.-i\\=1*-i\\=-i[/tex]
b. i^28=i
[tex]i^{28}\\=i^{4*7}\\=(i^4)^7\\=(1)^7\\=1[/tex]
c. i^14=1
[tex]i^{14}\\=i^{12}.i^2\\=(i^{4*3}).i^2\\=> i^2=-1\\=(i^4)^3.-1\\=(1)^3*-1\\=-1[/tex]
d. i^8=1
[tex]i^8\\=i^{4*2}\\=(i^4)^2\\=(1)^2\\=1[/tex]
e. i^5=i
[tex]i^5\\=i^{2*2*1}\\=i^2*i^2*i\\=(-1)(-1)(i)\\=i[/tex]
Hence,
The correct simplifications of powers of i are:
a. i^27=−i
d. i^8=1
e. i^5=i
Keywords: Imaginary units, Iota
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