Solving a Complex Polynomial Equation

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In summary, the problem is to solve (z^4)+1 = 0, with the attempt at a solution being to factor it into ((z^2)+i)*((z^2)-i)= 0 and then solve for z^2= i and z^2= -i using polar form. Thanks to the help from HallsofIvy, the problem can now be solved.
  • #1
wam_mi
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Homework Statement



Happy New Year, everybody!

Problem: Solve (z^4)+1 = 0


Homework Equations





The Attempt at a Solution



Attempt: (z^4)+1 = 0
((z^2)+i)*((z^2)-i) = 0

Could anyone help me to complete this question please?

Many Thanks!
 
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  • #2
HINT: Let [itex]z = e^{i\theta}[/itex]
 
  • #3
wam_mi said:

Homework Statement



Happy New Year, everybody!

Problem: Solve (z^4)+1 = 0


Homework Equations





The Attempt at a Solution



Attempt: (z^4)+1 = 0
((z^2)+i)*((z^2)-i) = 0

Could anyone help me to complete this question please?

Many Thanks!
So you know that z^2+ i= 0 and z^2- i= 0. Now you need to solve z^2= i and z^2= -i.

You can, as Hootenanny suggested, write i and -i in "polar form" and use the fact that
[tex]\left(re^{i\theta}\right)^{1/2}= r^{1/2} e^{i\theta/2}[/tex]
 
  • #4
HallsofIvy said:
So you know that z^2+ i= 0 and z^2- i= 0. Now you need to solve z^2= i and z^2= -i.

You can, as Hootenanny suggested, write i and -i in "polar form" and use the fact that
[tex]\left(re^{i\theta}\right)^{1/2}= r^{1/2} e^{i\theta/2}[/tex]



Thanks a lot for your help HallsofIvy. I think I can do it now, thank you!
 
  • #5
HallsofIvy said:
So you know that z^2+ i= 0 and z^2- i= 0. Now you need to solve z^2= i and z^2= -i.

You can, as Hootenanny suggested, write i and -i in "polar form" and use the fact that
[tex]\left(re^{i\theta}\right)^{1/2}= r^{1/2} e^{i\theta/2}[/tex]


Thanks for the hint!
 

What is a polynomial equation?

A polynomial equation is an equation that contains at least one variable and involves only the basic operations of addition, subtraction, multiplication, and non-negative integer exponents.

How do you solve a polynomial equation?

To solve a polynomial equation, you need to isolate the variable on one side of the equation and simplify the other side until you have a solution. This can be done by using algebraic methods such as factoring, the quadratic formula, or synthetic division.

What are the different types of polynomial equations?

The three main types of polynomial equations are linear, quadratic, and cubic equations. Linear equations have one variable raised to the first power, quadratic equations have one variable raised to the second power, and cubic equations have one variable raised to the third power.

What is the degree of a polynomial equation?

The degree of a polynomial equation is the highest exponent of the variable in the equation. For example, the degree of the polynomial equation 3x^2 + 5x + 2 is 2, since the variable is raised to the highest power of 2.

How many solutions can a polynomial equation have?

The number of solutions a polynomial equation can have depends on its degree. A linear equation has one solution, a quadratic equation can have two solutions, and a cubic equation can have three solutions. In general, a polynomial equation of degree n can have up to n solutions.

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