Answer:
Therefore
[tex]i^{110}=-1[/tex]
[tex]i^{92}=1[/tex]
[tex]i^{245}=i[/tex]
[tex]i^{347}=-i[/tex]
Step-by-step explanation:
Complex Number :
A complex number is a number that can be expressed in the form
[tex]a + bi[/tex]
where a and b are real numbers, and
' i ' is called an imaginary number.
The value of i is given as
[tex]i=\sqrt{-1}[/tex]
Squaring both the side we get
[tex]i^{2}=(\sqrt{-1})^{2}\\\therefore i^{2}=-1[/tex]
Therefore on substituting we get
For i^110
[tex]i^{110}=(i^{2})^{55}=(-1)^{55}=-1[/tex]
Similarly for i^92
[tex]i^{92}=(i^{2})^{46}=(-1)^{46}=1[/tex]
Similarly for i^245
[tex]i^{245}=(i^{2})^{122}\times i=(-1)^{122}\times i=1\times i =i[/tex]
Similarly for i^347
[tex]i^{347}=(i^{2})^{173}\times i=(-1)^{173}\times i=-1\times i =-i[/tex]
Therefore
[tex]i^{110}=-1[/tex]
[tex]i^{92}=1[/tex]
[tex]i^{245}=i[/tex]
[tex]i^{347}=-i[/tex]
