5 degree equation contains imaginary value?

In summary, @Buffu suggests that polar form can be used to simplify a complex equation. @Mark44 does not know how to use polar form, but he is interested in finding out how.
  • #1
Helly123
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Homework Statement



15_Mat_B_-_1.5.png

Homework Equations


polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution


i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?
 
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  • #2
Helly123 said:

Homework Statement



View attachment 205810

Homework Equations


polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution


i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?

Are you allowed to use polar form ?
 
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  • #3
Helly123 said:

Homework Statement



View attachment 205810

Homework Equations


polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

The Attempt at a Solution


i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

and it didn't work out. can anyone show me other way to solve it?
I didn't verify your work above, but you should replace i^2 by -1, and i^3 by -i, and so on.

This problem is much easier if you use @Buffu's suggestion, assuming you know about polar form for complex numbers.

BTW, your "relevant equation" isn't relevant, as it's wrong. With an exponent of 2 on the left side, the summation should be ##\sum_{k=0}^2 \binom 2 k a^{2-k} b^k##.
 
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  • #4
Mark44 said:
I didn't verify your work above, but you should replace i^2 by -1, and i^3 by -i, and so on.

This problem is much easier if you use @Buffu's suggestion, assuming you know about polar form for complex numbers.

BTW, your "relevant equation" isn't relevant, as it's wrong. With an exponent of 2 on the left side, the summation should be ##\sum_{k=0}^2 \binom 2 k a^{2-k} b^k##.
why should I replace that? i^2 = -1 ? i^3 = -i? why?

Buffu said:
Are you allowed to use polar form ?
yes, what exactly the purpose of using polar form?
w = 1/2 + ( (root3) / 2) i
r = 1, tetha = 60degrees??
 
  • #5
Helly123 said:
why should I replace that? i^2 = -1 ? i^3 = -i? why?
Because by doing this you can reduce the 6 terms you showed to only 2 terms.
Helly123 said:
yes, what exactly the purpose of using polar form?
w = 1/2 + ( (root3) / 2) i
r = 1, tetha = 60degrees??
Multiplication of complex numbers is much easier in polar form. If ##z_1 = r_1e^{i\theta_1}## and ##z_2 = r_2e^{i\theta_2}##, then ##z_1z_2= r_1r_2e^{i(\theta_1 + \theta_2)}##

By the way, this Greek letter -- ##\theta## -- is theta, not tetha.
 
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  • #6
Mark44 said:
Because by doing this you can reduce the 6 terms you showed to only 2 terms.

Multiplication of complex numbers is much easier in polar form. If ##z_1 = r_1e^{i\theta_1}## and ##z_2 = r_2e^{i\theta_2}##, then ##z_1z_2= r_1r_2e^{i(\theta_1 + \theta_2)}##

By the way, this Greek letter -- ##\theta## -- is theta, not tetha.
i see..
replacing that not changing the value, so i^2 equal to -1 ?
i^3 equal to -i? what theory is used?
so i^4 = (i^2)^2 = (-1)^2 = 1?
i^5 = i^4 . i = i
 
  • #7
Helly123 said:
i see..
replacing that not changing the value, so i^2 equal to -1 ?
i^3 equal to -i? what theory is used?
so i^4 = (i^2)^2 = (-1)^2 = 1?
i^5 = i^4 . i = i
Yes to all. ##i^2 = -1## by definition. ##i^3 = i^2 i = -1 i = -i##, and so on.
 
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  • #8
I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?
 
  • #9
Helly123 said:
I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?
Leaving the complex number in Cartesian (or rectangular) form will work, but it takes more work.

In polar form ##\omega = \frac 1 2 + \frac {\sqrt 3} 2 i = e^{i \pi/3}##. Note that because ##|\omega| = 1## the r factor in the polar form is just 1, and I don't show it.
##\omega^5 = 1^5 e^{i5 \pi/3}##, which represents a complex number whose angle is ##5\pi/3## and whose magnitude is (still) 1.This exponential polar form can be converted into trig polar form, because ##e^{i\theta} = \cos(\theta) + i\sin(\theta)##, due to Euler's Formula.

So all you have to do for this problem is to evaluate ##\cos(5\pi/3)## and ##\sin(5\pi/3)##.
 
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  • #10
z = re = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
z5 = r5ei5θ = ei5π/3 = e-iπ/3 = 1/2 - i√3/2
 
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  • #11
Helly123 said:
I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
@Mark44 do you know how too?

I think you should have used ##\omega^3 = -1##, then you only need to evaluate ##\omega^2## which is easier .
 
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  • #12
Buffu said:
I think you should have used ##\omega^3 = 1##, then you only need to evaluate ##\omega^2## which is easier .
How w = 1? Isnt w^2 = -1 by definition?
 
  • #15
Helly123 said:
You meant w^3 = 1?

No I mean ##\omega^3 = -1 \implies (-\omega)^3 = 1##. Look at the link in my previous you can see three cube roots of ##1##.
 
  • #16
Buffu said:
No I mean ##\omega^3 = -1 \implies (-\omega)^3 = 1##. Look at the link in my previous you can see three cube roots of ##1##.
Hmm on the link i saw the property is w^3 = 1
W^5 = w^2
W^5 = 1 + root3 i + 3/4 i^2
While i^2 = -1
= 1 + root3 i - 3/4
= -1/4 + 4root 3 i / 4

It didnt work out.
If w^3 = -1
Then w^5 = 1/4 - 4root 3 i/4
 
  • #17
mjc123 said:
z = re = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
z5 = r5ei5θ = ei5π/3 = e-iπ/3 = 1/2 - i√3/2

How can ei5π/3 = e-iπ/3 ? @mjc123
 
  • #18
Helly123 said:
Hmm on the link i saw the property is w^3 = 1
W^5 = w^2
W^5 = 1 + root3 i + 3/4 i^2
While i^2 = -1
= 1 + root3 i - 3/4
= -1/4 + 4root 3 i / 4

It didnt work out.
If w^3 = -1
Then w^5 = 1/4 - 4root 3 i/4

In that page ##\omega^\prime = \dfrac{- 1 \pm \sqrt{3} i}{2}, 1## whereas here ##\omega =\dfrac{+1 + \sqrt{3} i}{2} = -\dfrac{1 + \sqrt{3} i}{2} = -\omega^\prime##.

Helly123 said:
How can ei5π/3 = e-iπ/3 ? @mjc123

##\exp(5i\pi/3) = \cos 5\pi/3 + i \sin 5\pi/3 = 1/2 + -\sqrt{3}i/2 = \exp(-\pi i/3)##
 
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  • #19
Helly123 said:
How can ei5π/3 = e-iπ/3 ? @mjc123
Adding a multiple of 2π to the angle brings you back to the same point. So e = ei(θ+2nπ)
So ei5π/3 = ei(5π/3 -2π) = e-iπ/3. Draw the Argand diagram to see this.
 
  • #20
mjc123 said:
Adding a multiple of 2π to the angle brings you back to the same point. So e = ei(θ+2nπ)
So ei5π/3 = ei(5π/3 -2π) = e-iπ/3. Draw the Argand diagram to see this.
why do I have to do it? If it's still theta?
 
  • #21
Helly123 said:
How can ei5π/3 = e-iπ/3 ?

Helly123 said:
why do I have to do it? If it's still theta?
You're missing the point -- it's not "still theta." ##-\pi/3## and ##5\pi/3## are different angles but ##e^{-i\pi/3}## and ##e^{i5\pi/3}## represent the same complex numbers, with terminal points at the same location on the unit circle. Drawing the Argand diagram, as @mjc123 suggests, makes this clear. So no, you don't have to do this, but doing so can help you understand it better.

To my way of thinking, any time you can combine both an algebraic explanation with drawing, it makes things more understandable.
 
  • #22
how if my theory is like this
$$ e^{5phi/3} = e^{300degrees} = e^{360-300} = e^{60} = cos 60 + i sin 60 = 1/2 - i root3/2 ??
 
  • #23
Helly123 said:
how if my theory is like this
$$ e^{5phi/3} = e^{300degrees} = e^{360-300} = e^{60} = cos 60 + i sin 60 = 1/2 - i root3/2 ??

You cannot possibly mean what you wrote: ##e^{60}##, which is about ##1.142 \times 10^{26}##, and this is not even close to ##1/2 - i \sqrt{3}/2##.
 
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  • #24
You've missed out the i and written phi for pi, but
ei5π/3 = ei*300° = ei*(300°-360°) = ei(-60°) = cos(-60°) + i sin (-60°) = cos 60° - i sin 60°C = 1/2 - i√3/2
It's 300 - 360, not 360 - 300. You can add (or subtract) any whole multiple of 360° to your original angle, and get the same angle.
 
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  • #25
mjc123 said:
You've missed out the i and written phi for pi, but
ei5π/3 = ei*300° = ei*(300°-360°) = ei(-60°) = cos(-60°) + i sin (-60°) = cos 60° - i sin 60°C = 1/2 - i√3/2
It's 300 - 360, not 360 - 300. You can add (or subtract) any whole multiple of 360° to your original angle, and get the same angle.
Why 300-360, not 360-300 ?
 
  • #26
Helly123 said:
Why 300-360, not 360-300 ?
Because ##\frac {5\pi} 3## (same measure as 300°) is an angle in the 4th quadrant. 60° is in the 1st quadrant.
 
  • #27
Helly123 said:
Why 300-360, not 360-300 ?
What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!
 
  • #28
mjc123 said:
What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!
i try to do what you suggested before, i just get it. So -60 has same 'appearance ' as 300 degrees? So cos 300 = cos -60
 
  • #29
mjc123 said:
What did I write immediately after that?
Did you do what I suggested and draw the Argand diagram? That should make it clear. Adding or subtracting any integer multiple of 2π (360°) takes you round a full circle and comes back to the same point. Subtracting the angle from 2π is quite different!
so if the question find cos -60. Then cos -60= cos (360-300) ?
 
  • #30
Helly123 said:
So -60 has same 'appearance ' as 300 degrees? So cos 300 = cos -60
Yes.
Helly123 said:
so if the question find cos -60. Then cos -60= cos (360-300) ?
In this case it happens to be, but be careful. Sin -60° is not equal to sin (360°-300°). -60° is equivalent to 360°-60° = 300°, not to 360°-300° = 60°.
cos -θ = cos 2π-θ = cosθ. sin -θ = sin 2π-θ = -sinθ
Distinguish between making a full circle i.e. θ → 2nπ+θ, and reflecting about the x axis, i.e. θ → -θ = 2π-θ. They are not the same operation, and the latter does not give the same angle.
 
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FAQ: 5 degree equation contains imaginary value?

1. What is a 5 degree equation?

A 5 degree equation is a mathematical expression that contains a variable raised to the power of 5, such as x^5. These types of equations are also known as quintic equations.

2. What does it mean for a 5 degree equation to contain imaginary values?

When a 5 degree equation contains imaginary values, it means that the solutions to the equation involve the use of imaginary numbers, which are numbers that can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1). These solutions cannot be represented on a traditional number line.

3. How do you solve a 5 degree equation with imaginary values?

To solve a 5 degree equation with imaginary values, you can use the techniques of complex number algebra, which involves manipulating equations with imaginary numbers. You can also use the quadratic formula to solve for the roots of the equation.

4. Why do some 5 degree equations have imaginary values?

Some 5 degree equations have imaginary values because they do not have any real number solutions. This can occur when the equation has a complex root, or when the discriminant (b^2-4ac) of the quadratic formula is negative.

5. What are some real-world applications of 5 degree equations with imaginary values?

5 degree equations with imaginary values have many real-world applications, such as in physics, engineering, and economics. For example, they can be used to model the motion of a pendulum, the growth of a population, or the behavior of electrical circuits.

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