Complex Roots of Z^n = a + bi: Finding Solutions Using De Moivre's Theorem

[Wi_N]

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

 

[Logical Dog]

Shouldnt it be k = { 0 , 1, 2, 3} I used this formula to find roots: (edited)

$$80*e^{-i\frac{((3/4)\Pi) + 2\Pi*K }{4}}$$

 

[Wi_N]

Shouldnt it be k = { 0 , 1, 2, 3} I used this formula to find roots:

$$80*e^{\frac{-i((3/4)\Pi) + 2\Pi*K }{4}}$$

yes but first root already has k=0

 

[FactChecker]

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?

Yes. This looks ok to me. Is there some reason you think it is not?

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?

This also looks correct to me. Maybe I'm overlooking something. What makes you think that you are wrong?

Homework Equations

The Attempt at a Solution

For the homework format, you should separate your work into the appropriate sections, not just append empty sections.

 

[Wi_N]

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.

getting wrong answer from the machine...

 

[FactChecker]

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]

This looks wrong to me. 'i' should multiply the entire argument and the argument of -80i is 3π/2 or -π/2.

 

[Logical Dog]

This looks wrong to me. 'i' should multiply the entire argument and the argument of -80i is 3π/2 or -π/2.

yes, sorry. $$80*e^{-i\frac{((3/2)\Pi) + 2\Pi*K }{4}}$$

 

[Wi_N]

so in short my answer is correct?

 

[FactChecker]

so in short my answer is correct?

I think so. Maybe something went wrong when you checked it.

 

[SammyS]

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

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?

 

[Wi_N]

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?

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

edit: yes its its a mistake it shouldb k2pi/4. thanks. strange that i got a wrong answer for z1 though.

 

[FactChecker]

There are at least a couple of errors here.

Your z₁ looks good.

z₂ is not. How can you have a negative modulus?

It's just -z₁, which is another 4'th root.

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?

Good catch. That is a mistake.

 

[FactChecker]

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

But shouldn't it be 3pi/8 + k2pi/4 ?

 

[Wi_N]

But shouldn't it be 3pi/8 + k2pi/4 ?

yes realized that now thanks. but i didnt use any k value for z1 and i still got the wrong answer =(.

 

[FactChecker]

yes realized that now thanks. but i didnt use any k value for z1 and i still got the wrong answer =(.

You should re-check your check with the computer calculation. I'm suspicious of that.

 

[Wi_N]

You should re-check your check with the computer calculation. I'm suspicious of that.

edit: i feel like the biggest idiot on the planet. its 81 not 80...

thanks to everyone for their help-

 

[FactChecker]

edit: i feel like the biggest idiot on the planet. its 81 not 80...

thanks to everyone for their help-

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. :>)

 

[Logical Dog]

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):

$$z^{4}= -80i$$https://www.codecogs.com/latex/eqneditor.php

 

[Wi_N]

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):

$$z^{4}= -80i$$https://www.codecogs.com/latex/eqneditor.php

thanks will check it out.