Solving a Complex Equation: Express as e^(iθ)

In summary, the conversation discusses how to express a complex number in exponential form and how to solve an equation using this form. The solution involves multiplying by -1 in order to account for the given negative value, and the equation is simplified to a fifth-order equation with five roots. The conversation also touches on using algebra to determine the number of roots in an equation."
  • #1
icystrike
445
1

Homework Statement


Express the complex number in exp. form
[tex]-\frac{1}{2}(1+i\sqrt{3})[/tex]

Solve the following eqn:
[tex](w+2)^{4}=-\frac{1}{2}(1+i\sqrt{3})[/tex]

Homework Equations


The Attempt at a Solution


[tex]e^{i\frac{\pi}{3}}[/tex]

[tex]w+2=e^{(\frac{\frac{\pi}{3}+2k\pi}{4})i}
=e^{(\frac{\pi}{12}+\frac{\pi}{2}k)i}
[/tex]

Such that k=0,1,2,3
 
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  • #2
Hi icystrike! :smile:

Your first answer, eπi/3, would be the correct answer for +1/2 (1 + i√3).

So, to get minus that, multiply by … ? :wink:

For the second answer, the ekπi/2 can be simplified a little, for example by using a "±" (and of course, put the 2 on the other side).
 
  • #3
tiny-tim said:
Hi icystrike! :smile:

Your first answer, eπi/3, would be the correct answer for +1/2 (1 + i√3).

So, to get minus that, multiply by … ? :wink:

For the second answer, the ekπi/2 can be simplified a little, for example by using a "±" (and of course, put the 2 on the other side).

multiply by -1 . so it means that we have to keep a look out to or not multiply the ex. form by -1 right?
 
  • #4
uhh? :confused:

Hint: e? = -1 ? :smile:
 
  • #5
tiny-tim said:
uhh? :confused:

Hint: e? = -1 ? :smile:

multiply by [tex]e^{i\pi}[/tex] thus combine the power by law of indices
 
  • #6
Yup! … so instead of eiπ/3, it's … ? :smile:
 
  • #7
tiny-tim said:
Yup! … so instead of eiπ/3, it's … ? :smile:


[tex]e^{\frac{4\pi}{3}}[/tex]

yea?
 
  • #8
(just got up :zzz: …)

yea! :smile:
 
  • #9
Hi tiny-tim! Can you help me with this question?

Explain why the equation [tex](z+2i)^{6}=z^{6}[/tex] has five roots.

I thought it should be 6 roots?
 
  • #10
Hi icystrike! :smile:

erm :redface: … it's only a fifth-order equation! :rolleyes:
 
  • #11
hmm.. how do you tell? always thought that if we have power 6 , it will be 6 roots.
 
  • #12
You don't like algebra, do you? :redface:

Expand the LHS, and subtract the RHS … what happens? :smile:
 
  • #13

1. How do you express a complex equation as e^(iθ)?

To express a complex equation in the form of e^(iθ), you need to convert all the terms to their polar form. Then, you can use Euler's formula e^(iθ) = cos(θ) + isin(θ) to write the equation in the desired form.

2. What is the significance of e^(iθ) in solving complex equations?

e^(iθ) is a crucial component in solving complex equations because it allows us to represent complex numbers in a simplified form. It also helps in performing operations like multiplication, division, and powers on complex numbers more easily.

3. Can you use e^(iθ) to solve any complex equation?

Yes, e^(iθ) can be used to solve any complex equation. It is a powerful tool that simplifies the calculations and gives a better understanding of the solutions to complex problems.

4. How does e^(iθ) relate to the unit circle in trigonometry?

The value of e^(iθ) corresponds to a point on the unit circle in trigonometry, where θ is the angle formed by the point and the positive x-axis. This connection between e^(iθ) and the unit circle is the basis of Euler's formula.

5. Are there any other forms in which complex equations can be expressed?

Yes, complex equations can be expressed in different forms, such as rectangular form (a + bi), polar form (r(cosθ + isinθ)), and exponential form (re^(iθ)). However, e^(iθ) is the most commonly used form due to its simplicity and utility in solving complex equations.

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