Finding a complex number from an equation

In summary, the student attempted to solve a homework equation using the sum of cubes and the polar form, but failed to get a solution. After typing out the equation, they found their error.
  • #1
Temp0
79
0

Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.
 
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  • #2
Temp0 said:

Homework Statement



Find all complex numbers z such that z^6 = -64

Homework Equations



I tried it in two ways, once with the sum of cubes (a + b)(a^2 - ab + b^2)
as well as turning it into polar form and attempting it that way, z = re^(iθ)

The Attempt at a Solution



I completely couldn't solve it at all with the sum of cubes, however, I managed to get six solutions by using the polar form and then saying that θ = ∏/3 (3 + k), where k = 0, 1, 2, 3, 4, 5

By subbing in k into the equation z = r(cosθ + isinθ), where r = 2, I get the answers: -2i, +-(-1 - sqrt(3)i), +- (1 + sqrt(3)i), and +2i. However, the book I'm using gives the answers +- 2i, and +- (sqrt(3) + i), and +- (-sqrt(3) - i). Thank you guys in advance.

Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?
 
  • #3
We used to solve these in a geometrically way noting that the roots will be on a circle equally spaced at every 60 degrees about 0+0i and with a radius of 2. So given one root you can find the others.
 
Last edited:
  • #4
Dick said:
Well, (-1 - sqrt(3)i)^6=64, not -64. Something clearly went wrong with your polar solution. Can you show us what you did?

Well, r^6 = 64 so r = 2, and then because a / r = - 32, and b / r = 0, I just assumed that the angle I start at is pi, so my angle equation becomes 6θ = ∏ + 2∏k, and oh wait I pretty much just found my error as soon as I typed it out LOL, thanks alot.
 

1. How do you find the complex roots of an equation?

To find the complex roots of an equation, you can use the quadratic formula or the method of completing the square. You may also use a graphing calculator or software to visualize the complex roots.

2. What is a complex number?

A complex number is a number that has both a real and imaginary component. It is usually written in the form a + bi, where a is the real part and bi is the imaginary part. The imaginary part is represented by the letter i, which is the square root of -1.

3. Can you have more than one complex root for an equation?

Yes, an equation can have multiple complex roots. This occurs when the discriminant, b^2-4ac, is less than 0. In this case, the roots will be complex conjugates of each other.

4. How do you solve equations with complex numbers in them?

To solve equations with complex numbers, you can use the same techniques as solving equations with real numbers. You can combine like terms, isolate the variable, and use inverse operations. But when dealing with complex numbers, you must also keep in mind the properties of imaginary numbers, such as i^2 = -1.

5. Can complex numbers be used in real-life applications?

Yes, complex numbers are used in various fields, including physics, engineering, and finance. They can represent quantities that have both magnitude and direction, such as alternating current in electrical circuits. They are also used in signal processing and control systems.

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