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

In summary, the homework statement asks for the value of (-√3 + i)43/243. The attempt at a solution states that i0=1, i1=i, i2=-1, i3=-i, and that he tried to use that to help but he got no where. He also tried to break up the exponent into smaller numbers however that did not work for him either. He is looking for help to better understand the problem.
  • #1
ver_mathstats
260
21

Homework Statement


Find the value of (-√3 + i)43/243

Homework Equations

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.
 
Physics news on Phys.org
  • #3
ver_mathstats said:

Homework Statement


Find the value of (-√3 + i)43/243

Homework Equations

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.
 
  • #4
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 ...
 
  • #5
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.
 
  • Like
Likes FactChecker
  • #6
Ray Vickson said:
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.
 
  • #7
Orodruin said:
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?
 
  • #8
DaveE said:
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.
 
  • #9
Orodruin said:
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?
 
  • Like
Likes DaveE
  • #10
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:
  • #11
ver_mathstats said:
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.##
 
  • Like
Likes FactChecker

1. How do I find the value of a complex number with a high exponent?

To find the value of a complex number with a high exponent, you can use the polar form of the complex number. This involves converting the complex number into its magnitude and angle form, raising it to the desired exponent, and then converting it back to rectangular form. This method is known as De Moivre's theorem.

2. Can I use a calculator to find the value of a complex number with a high exponent?

Yes, most scientific calculators have a function for calculating complex numbers with high exponents. You can use the "power" or "exponent" button and enter the complex number and exponent to get the result. However, it is important to make sure your calculator is set to the correct mode (polar or rectangular) before performing the calculation.

3. What is the difference between using polar and rectangular forms to find the value of a complex number with a high exponent?

The polar form involves converting the complex number into its magnitude and angle form, which can be easier to work with when dealing with high exponents. The rectangular form, on the other hand, is the standard form of representing complex numbers and may be more familiar to some individuals. Both methods will yield the same result.

4. Are there any special cases when finding the value of a complex number with a high exponent?

Yes, when the exponent is a multiple of 4 (4, 8, 12, etc.), the result will always be a real number. This is because the angle of the complex number will be a multiple of 90 degrees, resulting in a real number when raised to a power. In all other cases, the result will be a complex number.

5. Can I use De Moivre's theorem to find the value of a complex number with a non-integer exponent?

Yes, De Moivre's theorem can be extended to work with non-integer exponents. In this case, the complex number will be raised to a power using the binomial theorem. The resulting answer will be a complex number in polar form, which can then be converted back to rectangular form if needed.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
859
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
167
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • General Math
Replies
7
Views
1K
Back
Top