Iotic iota?

We can write the given expression as $i^{2} + (i^{2})^{3} + (i^{2})^{2}.$ Since $i^{2} = - 1$ , we substitute it in the above expression. Therefore $\begin{array}{c}& \ \ (-1) + (-1)^{3} + (-1)^{2} \\ & =-1 + (-1) + 1 \\ & =-1. \square \end{array}$

$i=\sqrt{-1}$

So $i^2=-1$ .

$\Rightarrow i^6=(i^2)^3=-1$

And $i^4=1$ .

So the ans is -1. $\square$

Since $i = \sqrt{-1},$ then $i^2 = -1$ ,

then $i^6 = (i^2)^3 = (-1)^3 = -1$

and $i^4 = (i^2)^2 = (-1)^2 = 1$ .

Therefore:

$i^2 + i^6 + i^4 = -1 + (-1) + (1) = -2 + 1 = -1$

i^2 = -1 therefore: i^2 + (i^2)(i^2)(i^2) + (i^2)(i^2) = -1+(-1)(-1)(-1) +(-1)(-1) = -1 ans...

(-1)^2+(-1)^6+(-1)^=-1+(-1)+1=-1

i is an sqrt of -1. i^2 is -1, i^6 is {sqrt(-1)}^6 so (-1)^3. It is -1, i^4 is {sqrt(-1)}^4 So (-1)^2. It is 1. In total, -1-1+1 is -1.

i^2 + i^6 + i^4 = (i^^2) (1 + i^4 + i^2) = (-1) (1 + 1 + (-1)) = (-1)*(1) = -1

Same solution as @Bilal Khan

$i^2+i^6+i^4=i^2+(i^2)^3+(i^2)^2=-1+(-1)^3+(-1)^2=-1+-1+1=\boxed{\large{-1}}$

As we know, i^2=-1. i^3 can be written as i(-1)=-i i^4 is (-i)i=1 i^5 is 1(i) = i i^6 is i(i) which is also i^2=-1

So (i^2)+(i^6)+(i^4) is -1 + -1 + 1 = -1

i^{2} + i^{6} + i^{4}

i^{2} = (√-1)^{2} = -1

i^{6} = (√-1)^{6} = (-1)^{3} = -1

i^{4} = (√-1)^{4} = (-1)^{2} = 1

-1-1+1 = -1

-1+-1 +1 =-2+1 = -1

(i^2= -1+ i^6= -1+ i^4= 1)= -1

i^2+i^6+i^4 = i^2(1+i^4+i^2) = -1(1+i^4-1) = -(i^2)^2 = -(-1)^2 = -1*1 = -1

since i^2=-1 so -1-1+1=-1

as i=-1^1/2 i2=-1 i6 =-1 i4=1 so the sum is -1

i^2+i^6+i^4 = -1-1+1 = -1

i^{2} = (-1) hence i^{6} = (-1)^{3} = -1 i^{4} = (-1)^{2} = +1 there fore (-1) + (-1) + (+1) = (-1)

i^2=-1, i^6=-1, i^4=1 ----> (-1)+(-1)+1=-1

= i^2 ( 1 + i^4 + i^2 )

= i^2 { 1 + (i^2)^2 + i^2 }

= i^2 { 1 + (-1)^2 + (-1) } ; [ i^2 = -1]

= i^2 ( 1+ 1 - 1 )

= i^2

= -1 Ans.

i = root -1 i^2 = -1 i^3 = -root - 1 i^4 = 1 This is an ongoing cycle so i^5 goes back to the same value as (i) since it is i^4 x i = 1 x i = i. also, i^6 = i^2, i^7 = i^3, i^8 = i^4 i^9 = i and this is a cycle that goes on forever so this equation will simplify to: -1 -1 + 1 = -1

-1-1+(1)=-1

As i² = -1 , i^4 = (i^2)^2 = (-1)^2 = 1 , i^6 = (i^2)^3 = (-1)^3 = -1

Therefore , -1+(-1) + 1 = -1

since i^2 = -1

therefore i^2 = -1 and i^6 = -1 and i^4 = 1

therefore i^2 + i^6 + i^4 = -1-1+ = -1

i^2+i^4(i^2+1)=i^2+0= -1

ι^2 = -1

ι^2+ι^4+ι^6=(−1)+(−1)^3+(−1)^2

= -1 -1 +1

=-1

i=root(-1) so this way value is -1