Question 1208825: Write each expression in the standard form a + bi.
1. i^(-23)
2. (1 + i^3
Found 3 solutions by math_tutor2020, ikleyn, Edwin McCravy:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'll focus on problem #1 only.
Problem #2 seems to have a typo in it. I have a feeling it's supposed to say (1+i)^3, but I'm not sure.
Anyway, with problem #1, let's consider i^23 and worry about the negative exponent later.
To evaluate i^23, we divide the exponent by 4 and look at the remainder.
Refer to this similar question to see why we divide by 4.
23/4 = 5 remainder 3
This means, i^23 = i^3 = -i
Another way we can arrive at that is to say:
i^(23) = i^(20+3)
= i^(20)*i^3
= i^(5*4)*i^3
= (i^4)^5*i^3
= (1)^5*i^3
= i^3
= -i
Notice I rewrote 23 as 20+3. The 20 is the largest multiple of 4 just short of 23.
The useful exponent rules are a^b*a^c = a^(b+c) and (a^b)^c = a^(b*c).
There are probably other ways of determining that i^23 = i^3 = -i.
Now to account for the negative exponent, we could have these steps
i^(-23) = 1/(i^23)
= 1/(-i)
= i/(-i^2)
= i/(-(-1))
= i
The jump from the 2nd line to the 3rd line is when we multiply top/bottom by i.
Doing this step will convert the imaginary denominator to a real number.
Answer: i^(-23) = i
Verification with WolframAlpha
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
In your expression (2), there is an opening parenthesis, but there is no closing parenthesis,
so, the expression is heavily sick.
Answer by Edwin McCravy(20064)  (Show Source):
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