evaluate i14^

[itsmelana]

thank you!

 

[tkhunny]

Do you mean "i raised to the 14th power"?

You need only this information:

\(\displaystyle i^{1} = i\)

\(\displaystyle i^{2} = -1\)

\(\displaystyle i^{3} = i*i^{2} = -i\)

\(\displaystyle i^{4} = i^{2}*i^{2} = (-1)(-1) = 1\)

You just have to figure out where 14 comes in this sequence.

 

[itsmelana]

yes, i to the 14th power

 

[pka]

If \(\displaystyle n = 4k + r\,,\,0 \le r < 4\)
then \(\displaystyle i^n = i^{4k + r} = \left( {i^{4k} } \right)\left( {i^r } \right) = \left( {i^4 } \right)^k \left( {i^r } \right) = \left( 1 \right)^k \left( {i^r } \right) = \left( {i^r } \right)\)
\(\displaystyle 14=3(4)+2\)

 

[masters]

yes, i to the 14th power

Divide the exponent by 4

If the remainder is 0, the answer is 1

If the remainder is 1, the answer is i

If the remainder is 2, the answer is -1

If the remainder is 3, the answer is -i

14/4 = 3 remainder 2. Therefore i^14 = -1

 

[pka]

masters,
Did you bother yourself to read my response to this question?
That will work for any integer n.
Will your approach? Say n=-23?

 

[masters]

Re: evaluate i^14

masters,
Did you bother yourself to read my response to this question?Yes, I did bother myself to read your post.
That will work for any integer n.I did assume a positive exponent because that was what was presented. I only presented a simplistic way to determine any positive power of i.

Will your approach? Say n=-23?Can you appreciate my approach to this specific question? See below.

Sorry, lost my head. I was just following up on Tkhunny's post. I'm so embarrassed.

But I believe that \(\displaystyle i^{-23}=\frac{1}{i^{23}}\)

\(\displaystyle 23\div4=5 \ \ remainder \ \ 3\)

\(\displaystyle i^{-23}=\frac{1}{i^{23}}=\frac{1}{-i}=\frac{1\cdot -i}{-i \cdot -i}=\frac{-i}{-1}=i\)

Still works in principle. You just have to remember to use the reciprocal since you made the exponent negative and then rationalize the denominator if necessary.