# What is the value of the number  i^231 ? Is it real or complex number?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.