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

[ver_mathstats]

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.

 

[Orodruin]

Polar form.

 

[Ray Vickson]

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.

 

[Orodruin]

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 ...

 

[DaveE]

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*e^di ). 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.

 

[ver_mathstats]

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.

 

[ver_mathstats]

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?

 

[ver_mathstats]

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*e^di ). 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.

 

[ver_mathstats]

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]

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.

 

[Ray Vickson]

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 $z^{43} = e^{i\: 36 \pi} e^{-i \pi/6} = e^{-i \pi/6} = - z.$