What is the value of the number i^231 ? Is it real or complex number?
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calendarEducator since 2010
write12,544 answers
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We have to determine if i^231 is real or complex.
Now i^2 = -1 , i^4 = 1
i^231 = i^228 * i^3
=> i^228* i^2*i
=> 1*-1*i
=> -i
Therefore i^231 is complex and equal to -i.
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Given the number i^231
Let us simplify and determine the values.
First we will rewrite the power.
==> i^(231) = i^(3+228)
Now we will use exponent properties to simplify.
We know that x^(a+b) = x^a * x^b
==> i^(3+228) = i^3 * i^228
But we know that i^2 = -1 ==> i^3 = -1*i = -i
==> i^231 = -i * (i^228)
Now we will rewrite the power again.
==> i^231 = -i * i^(2*114)
From exponent properties we know that x^ab = x^a^b
==> i^231 = -i *( i^2)^114
But i^2 = -1
==> i^231 = -i * (-1)^114
==> i^231 = -i * 1 = -i
==> i^231 = -i, and it is a complex number (-sqrt-1)
Q:What is the value of the number i^231 ? Is it real orcomplex number (edited please).
A:
We take the help of D' Moivre's theorem which states that (cos x+isin x)^n = cos nx+isin nx.
Here i = 0+i*1.
=> i = cos pi/2 + isin pi/2.
=> i^231 = cos(p/2i+isinpi/2)^231
=> i^231 = cos(231/2)pi + i*sin (231pi/2) by D' Moivre's theorem
=> i6231 = cos(114pi +3/2)pi+ i sin (114+3)pi.
=> i^231 = cos 3pi/2+ isin 3pi/2. as cosx+isinx is 2pi periodic.
=> i^231 = 0 +i*(-1)
=> i^231 = - i.
Therefore i^231 = -i which is complex number.
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