SOLUTION: Write each expression in the standard form a+bi. How do I simplify i^n when n is a big number? Two samples: A. i^14 B. 2i^(8)(2 + i^4)

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Question 1207482: Write each expression in the standard form a+bi.

How do I simplify i^n when n is a big number?

Two samples:

A. i^14

B. 2i^(8)(2 + i^4)


Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

When dealing with large numbers, try to look at small examples to see if a pattern can be formed.

We'll look at the first few powers of i
i^0 = 1
i^1 = i
i^2 = ( sqrt(-1) )^2 = -1
i^3 = i*i^2 = i*(-1) = -i
i^4 = ( i^2 )^2 = (-1)^2 = 1
i^5 = i*i^4 = i*1 = i
Etc

The summarized list would be
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
Etc

Be sure not to mix up the lowercase "eye" with the numeric digit "one".
The two symbols unfortunately look similar at a quick glance.

Anyways, the pattern we found was: 1, i, -1, -i, 1, i, ...
As you can see, once reaching 1, the pattern starts over again.
Therefore, this sequence repeats itself every 4 elements.

Why is this useful? Because we can divide the large-ish exponent of 14 by 4 to determine the remainder.
14/4 = 3 remainder 2
The "remainder 2" leads us to i^2 since it has exponent of 2.
Ignore the quotient. We only care about the remainder.

i^14 = i^2 = -1 which is the answer to part A.

Another example:
i^15 = i^3 = -i
since 15/4 gives some quotient with remainder 3

Another example:
i^98 = i^2 = -1
since 98/4 yields remainder 2.


If in doubt, turn to something like WolframAlpha to help confirm the answer
https://www.wolframalpha.com/input/?i=i%5E14
I use this tool a lot. Another tool that I use often is GeoGebra.
There are many other similar tools on the web which I encourage you to explore your favorite.


I'll leave part B for the student to handle.
If you are still stuck, or have any questions, then please let me know.
Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

The number of "i" is very special.


Consider the sequence i, i^2, i^3, i^4, i^5, i^6, i^7, i^8, . . . 


This sequence is      i, -1,  -i,   1,   i,   -1,  -i,  1,  . . . 


Thus the sequence is cyclic and the numbers in it repeat with the period of 4.


Therefore, if you need to find  i^n,  you should find the remainder  m = (n mod 4)  
of the index n modulo 4.

Then  i^n = i^m, which is one of four numbers i, -1, -i or 1.



For example, for i^14,  the index is  n= 14;  the remainder  (14 mod 4) is 2;
                        therefore,  i^14 = i^2 = -1.


Next example, for i^73, the index is  n= 73;  the remainder  (73 mod 4) is 1;
                        therefore,  i^73 = i^1 = i.


Thus, ALL calculations of this kind are quick, simple and easy. Making them is a  fun.



Regarding your question (B),  i^8 = 1;  i^4 = 1;  so, the value of (B) is  2*1*(2+1) = 2*3 = 6.

Solved.



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