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How to find the value of a complex number with high exponent

1. The problem statement, all variables and given/known data
Find the value of (-√3 + i)43/243

2. Relevant equations


3. The attempt at a solution
I do not know how to really go about this problem.

I know that i0=1, i1=i, i2=-1, i3=-i, and I tried to use that to help but I got to no where, I also tried to break up the exponent into smaller numbers however that did not work for me either. So I could use some help to better understand.

Thank you.
 

Ray Vickson

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1. The problem statement, all variables and given/known data
Find the value of (-√3 + i)43/243

2. Relevant equations


3. The attempt at a solution
I do not know how to really go about this problem.

I know that i0=1, i1=i, i2=-1, i3=-i, and I tried to use that to help but I got to no where, I also tried to break up the exponent into smaller numbers however that did not work for me either. So I could use some help to better understand.

Thank you.
Instead of converting to polar form, you could also do it directly, using a binomial expansion:
$$\text{Answer} = \sum_{k=0}^{43} {43 \choose k} (-1)^k \, 3^{k/2} \, i^{43-k}$$
Unless you have a lot of spare time I would not recommend you do this manually, but it is actually the method used by some computer algebra systems.
 

Orodruin

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Yet another method to minimize actual computations if not going to polar form:
  • Compute ##z^2##, then ##z^4 = z^2 z^2## using the result, then ##z^8 = z^4z^4##, etc
  • Write ##z^{43} = z^{32}z^{11} = z^{32}z^8z^3 = z^{32}z^8 z^2 z##.
In your case you could also simplify a bit by computing a few powers of ##z## just by repeated multiplication. Try multiplying ##z^2##, ##z^3##, ##z^4## for some time to see if you get something nicer ...
 
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A general comment about complex notation:
In math class complex numbers are often written in the rectangular form ( a + bi ), This is good for addition and subtraction calculations.
However, out in the real world where people work with complex numbers, they almost always use the polar form ( c*edi ). The polar form is much easier to work with for multiplication, exponents, etc. I think it's also easier to understand the physical interpretations of polar form, but that may just be a personal preference.
 
Instead of converting to polar form, you could also do it directly, using a binomial expansion:
$$\text{Answer} = \sum_{k=0}^{43} {43 \choose k} (-1)^k \, 3^{k/2} \, i^{43-k}$$
Unless you have a lot of spare time I would not recommend you do this manually, but it is actually the method used by some computer algebra systems.
Yes thank you, I was planning on trying that.
 
Yet another method to minimize actual computations if not going to polar form:
  • Compute ##z^2##, then ##z^4 = z^2 z^2## using the result, then ##z^8 = z^4z^4##, etc
  • Write ##z^{43} = z^{32}z^{11} = z^{32}z^8z^3 = z^{32}z^8 z^2 z##.
In your case you could also simplify a bit by computing a few powers of ##z## just by repeated multiplication. Try multiplying ##z^2##, ##z^3##, ##z^4## for some time to see if you get something nicer ...
Thank you. That's what I was thinking of doing since the homework question wants us to answer in the form a+bi, however when we do that we would change the exponent for both the numerator and denominator correct?
 
A general comment about complex notation:
In math class complex numbers are often written in the rectangular form ( a + bi ), This is good for addition and subtraction calculations.
However, out in the real world where people work with complex numbers, they almost always use the polar form ( c*edi ). The polar form is much easier to work with for multiplication, exponents, etc. I think it's also easier to understand the physical interpretations of polar form, but that may just be a personal preference.
Thank you for your response. In our math class we are almost always required to answer in the form a + bi.
 
Yet another method to minimize actual computations if not going to polar form:
  • Compute ##z^2##, then ##z^4 = z^2 z^2## using the result, then ##z^8 = z^4z^4##, etc
  • Write ##z^{43} = z^{32}z^{11} = z^{32}z^8z^3 = z^{32}z^8 z^2 z##.
In your case you could also simplify a bit by computing a few powers of ##z## just by repeated multiplication. Try multiplying ##z^2##, ##z^3##, ##z^4## for some time to see if you get something nicer ...
Would the correct answer be √3/2-1/2i?
 

FactChecker

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Convert to polar form. Then the power is easy to calculate to an answer in polar form. Convert the answer back to the a+ib form if you think that is necessary. That approach is almost certainly what is expected in this problem and will apply the most useful, fundamental facts about complex numbers.
 
Last edited:

Ray Vickson

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Thank you for your response. In our math class we are almost always required to answer in the form a + bi.
You want ##z^{43}##, where ##z = \frac{-\sqrt{3}} 2 + \frac1 2 i = e^{i \: (5/6) \pi}.## The answer will have the form
$$z^{43} = e^{i \: (43)\frac 5 6 \pi} = e^{i \: (215/6) \pi}$$
But ##215/6 = 36 - \frac 1 6,## so ##z^{43} = e^{i\: 36 \pi} e^{-i \pi/6} = e^{-i \pi/6} = - z.##
 

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