Bipolar Demon said:Shouldnt it be k = { 0 , 1, 2, 3} I used this formula to find roots:
[tex] 80*e^{\frac{-i((3/4)\Pi) + 2\Pi*K }{4}} [/tex]
Yes. This looks ok to me. Is there some reason you think it is not?Wi_N said:Homework Statement
I just can't seem to get the right answer. z^4+80i=0
looking at the complex plane u see the radius=r=80 (obviously)
using De Moivre extension: z^n=(r^(1/n))(cos((x/n)+k2pi/n)-isin((x/n)+k2pi/n)z1=((80)^(1/4))(cos(3pi/8)+isin(3pi/8)
shouldnt this be a root?
This also looks correct to me. Maybe I'm overlooking something. What makes you think that you are wrong?z2= -((80)^(1/4))(cos(3pi/8)+isin(3pi/8)
z3=((80)^(1/4))(cos((3pi/8)+(k2pi/8)))+isin((3pi/8)+(k2pi/8))
for k=1, 2.
what am i doing wrong?
For the homework format, you should separate your work into the appropriate sections, not just append empty sections.Homework Equations
The Attempt at a Solution
FactChecker said:Yes. This looks ok to me. Is there some reason you think it is not?This also looks correct to me. Maybe I'm overlooking something. What makes you think that you are wrong?For the homework format, you should separate your work into the appropriate sections, not just append empty sections.
This looks wrong to me. 'i' should multiply the entire argument and the argument of -80i is 3π/2 or -π/2.Bipolar Demon said:Shouldnt it be k = { 0 , 1, 2, 3} I used this formula to find roots:
[tex] 80*e^{\frac{-i((3/4)\Pi) + 2\Pi*K }{4}} [/tex]
FactChecker said:This looks wrong to me. 'i' should multiply the entire argument and the argument of -80i is 3π/2 or -π/2.
I think so. Maybe something went wrong when you checked it.Wi_N said:so in short my answer is correct?
There are at least a couple of errors here.Wi_N said:Homework Statement
I just can't seem to get the right answer. z^4+80i=0
looking at the complex plane u see the radius=r=80 (obviously)
using De Moivre extension: z^n=(r^(1/n))(cos((x/n)+k2pi/n)-isin((x/n)+k2pi/n)z1=((80)^(1/4))(cos(3pi/8)+isin(3pi/8)
shouldnt this be a root?z2= -((80)^(1/4))(cos(3pi/8)+isin(3pi/8)
z3=((80)^(1/4))(cos((3pi/8)+(k2pi/8)))+isin((3pi/8)+(k2pi/8))
for k=1, 2.
what am i doing wrong?
Homework Equations
The Attempt at a Solution
SammyS said:There are at least a couple of errors here.
Your z1 looks good.
z2 is not. How can you have a negative modulus?
Write ## -80i\ ## in terms of a negative angle, and work with that, (if that's what you intended by the negative).
For z3 (& z4 ?): Why are you dividing 2π by 8 ? It's not an 8th root that you want, is it?
It's just -z1, which is another 4'th root.SammyS said:There are at least a couple of errors here.
Your z1 looks good.
z2 is not. How can you have a negative modulus?
Good catch. That is a mistake.Write ## -80i\ ## in terms of a negative angle, and work with that, (if that's what you intended by the negative).
For z3 (& z4 ?): Why are you dividing 2π by 8 ? It's not an 8th root that you want, is it?
But shouldn't it be 3pi/8 + k2pi/4 ?Wi_N said:because when z^n and n is an even number the first roots are +-
if z1 is good then i don't know what the problem is. as for z3 z4 i probably wrote them wrong. point is that the argument is (3pi/2)/n in this case n=4
so 3pi/8 + k2pi/8
FactChecker said:But shouldn't it be 3pi/8 + k2pi/4 ?
You should re-check your check with the computer calculation. I'm suspicious of that.Wi_N said:yes realized that now thanks. but i didnt use any k value for z1 and i still got the wrong answer =(.
FactChecker said:You should re-check your check with the computer calculation. I'm suspicious of that.
No problem. The error that @SammyS spotted was important to find anyway. If you didn't make the 80 versus 81 mistake, that probably would have never been found. :>)Wi_N said:edit: i feel like the biggest idiot on the planet. its 81 not 80...
thanks to everyone for their help-
Bipolar Demon said:hello OP can i sugges you use latex editor online, I am learning it but this is much quicker and your work looks cleaner (so easy to spot any errors):
[tex] z^{4}= -80i [/tex]https://www.codecogs.com/latex/eqneditor.php
The value of Z in the equation Z^4=-80i is 2i. This can be found by taking the fourth root of -80i, which is equivalent to multiplying the fourth root of -1 by the fourth root of 80i. The fourth root of -1 is 1i, and the fourth root of 80i is 2i.
To solve for Z in the equation Z^n=a+bi, you can use the polar form of complex numbers. First, convert the complex number a+bi to polar form by finding its magnitude (r) and angle (θ). Then, use the formula Z=r^(1/n)*(cos(θ/n)+i*sin(θ/n)) to find the n-th root of Z. Finally, convert the result back to rectangular form by using the formula a+bi=r*cos(θ)+i*r*sin(θ).
Yes, Z^n can equal a complex number. In fact, in the given equation Z^n=a+bi, a+bi is a complex number. Z can also be a complex number, depending on the values of n, a, and b. The solution for Z will be a complex number if n is an even number or if n is a complex number itself.
The variable n in the equation Z^n=a+bi represents the power or exponent of Z. It determines the number of times Z is multiplied by itself. The solution for Z will vary depending on the value of n, with different values for n resulting in different solutions for Z.
The equation Z^n=a+bi can be represented on the complex plane. The complex number a+bi corresponds to a point on the plane, with a being the x-coordinate and b being the y-coordinate. The value of n determines the number of times Z (represented by a vector from the origin to the point a+bi) is rotated counterclockwise by the angle θ. The solution for Z can be found by finding the point on the complex plane that corresponds to the result of the rotation.