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# Help.

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Evaluate $i^{11} + i^{16} + i^{21} + i^{26} + i^{31}$.

This is an AoPS question but I don’t know what to do. Can someone help me understand? Just wanted to make that clear.

Jun 20, 2019

#1
+109345
+2

Note

i^2  = -1

And -1 raised to an even power = 1

And -1 raised to an odd power  =  - 1

i^11 =  I^10 * i     =   (i^2)^5 * i      =  (-1)^5 * i   = (-1) * i  =  - i

i^16  = (i^2)^8   = (-1)^8  =  1

i^21  = i^20 * i   =  (i^2)^10 * i  =  (-1)^10 * i   = 1 * i   =   i

i^26  = (i^2)^13  =  (-1)^13  = - 1

i^31  =  i^30 * i  =  (i^2)^15 * i   = (-1)^15 * i  =  -1 * i  = - i

So....adding the 5  results we get

-i + 1+ i - 1 - i  =

- i

Jun 20, 2019
edited by CPhill  Jun 20, 2019
#2
+91
+2

So

i^1 = i,

i ^ 2 = -1,

i^3 = -i,

i^4= 1,

i ^ 5 = i,

and it keeps repeating between the four (i,-1,-i,i).

So, to find i^11, we find 11mod4 (or the remainder when dividing 11 and 4) to get 3. So, i ^11 is -i.

Then you do the rest.

i^16 = 1

i^21 = i

i^26 = -1

i^31 = -i.

So, all of those added together is -i + 1 + i - 1 - i which is -i.

Jun 20, 2019
#3
+109345
0

THX, Pushy....I left out one....edit made!!!!!

CPhill  Jun 20, 2019
edited by CPhill  Jun 20, 2019
#4
+24388
+2

Evaluate

$$i^{11} + i^{16} + i^{21} + i^{26} + i^{31}$$

Formula: $$i^{4n+a} = i^a$$

$$\begin{array}{|rcll|} \hline &&\mathbf{ i^{11} + i^{16} + i^{21} + i^{26} + i^{31} } \\ &=& i^{4\cdot 2+3} + i^{4\cdot 4+0} + i^{4\cdot 5+1} + i^{4\cdot 6+2} + i^{4\cdot 7+3} \\ &=& i^{3} + i^{0} + i^{1} + i^{2} + i^{3} \\ &=& i^{2}i + 1 + i + i^{2} + i^{2}i \quad | \quad i^2 = -1 \\ &=& -i + 1 + i -1 + -i \\ &=& \mathbf{ -i } \\ \hline \end{array}$$

Jun 21, 2019