Complex numbers, solve equation

**fgg** · March 12th 2011, 01:18 AM

Hi,

I wonder how to solve this complex math problem with algebra:

z^4+z^2+1=0

Regards
/fgg

**mr fantastic** · March 12th 2011, 01:19 AM

Originally Posted by fgg

Hi,

I wonder how to solve this complex math problem with algebra:

z^4+z^2+1=0

Regards
/fgg

Start by substituting u = z^2 and solve the resulting quadratic equation for u. If you need more help, please show all your work and say where you get stuck.

**fgg** · March 12th 2011, 03:55 AM

Originally Posted by mr fantastic

Start by substituting u = z^2 and solve the resulting quadratic equation for u. If you need more help, please show all your work and say where you get stuck.

(z^2+(1/2))^2+1=0 after this I'm not sure how to preceed.

Other solution:

z^4=u^2

z^2+z+1=0

z=-(1/2)+-(3i)/2 z= - (1/2)+(3i)/2 z= - (1/2)+(3i)/2 no reell solution

z^2=The square root of - (1/2)+(3i)/2 and now ...?

**HallsofIvy** · March 12th 2011, 07:30 AM

Originally Posted by fgg

(z^2+(1/2))^2+1=0 after this I'm not sure how to preceed.

(z^2+ 1/2)^2= -1 and take the square root of both sides.,

Other solution:

z^4=u^2

z^2+z+1=0

z=-(1/2)+-(3i)/2 z= - (1/2)+(3i)/2 z= - (1/2)+(3i)/2 no reell solution

Well, you did title this thread "Complex numbers"!

z^2=The square root of - (1/2)+(3i)/2 and now ...?

Write -(1/2)+ (3/2)i in "polar form", $re^{i\theta}$ or $r(cos(\theta)+ i sin(\theta))$ , and use DeMoivre's formula.

**mr fantastic** · March 12th 2011, 12:43 PM

Originally Posted by fgg

(z^2+(1/2))^2+1=0 after this I'm not sure how to preceed. Mr F says: This is wrong.

Other solution:

z^4=u^2

z^2+z+1=0

z=-(1/2)+-(3i)/2 z= - (1/2)+(3i)/2 z= - (1/2)+(3i)/2 no reell solution

z^2=The square root of - (1/2)+(3i)/2 and now ...?

$\displaystyle u^2 + u + 1 = 0 \Rightarrow \left(u + \frac{1}{2}\right)^2 + \frac{3}{4} = 0$ .

But personally, I'd use the quadratic formula to solve for u and hence z^2. Knowing z^2 you can solve for z. If you need more help, please show all your work and say where you get stuck.

Your comment of "no reell (sic) solution" does not fill me with optimism that you have sufficient understanding of the course material to be attempting this question ....

**fgg** · March 12th 2011, 01:12 PM

Originally Posted by mr fantastic

$\displaystyle u^2 + u + 1 = 0 \Rightarrow \left(u + \frac{1}{2}\right)^2 + \frac{3}{4} = 0$ .

But personally, I'd use the quadratic formula to solve for u and hence z^2. Knowing z^2 you can solve for z. If you need more help, please show all your work and say where you get stuck.

Your comment of "no reell (sic) solution" does not fill me with optimism that you have sufficient understanding of the course material to be attempting this question ....

Well my English may not be the best since I'm from Scandinavia, sorry about that. What I meant by not a "reell" number is that I can not take the square root of a negative number since z is a real number.

After using Moivres formula I will have
z=(cos 178,9 + isin 178,9) but if I have to answer in the form z=x + yi
How would you then proceed?

Thank you for helping me

**mr fantastic** · March 12th 2011, 01:27 PM

Originally Posted by fgg

Well my English may not be the best since I'm from Scandinavia, sorry about that. What I meant by not a "reell" number is that I can not take the square root of a negative number since z is a real number.

After using Moivres formula I will have
z=(cos 178,9 + isin 178,9) but if I have to answer in the form z=x + yi
How would you then proceed?

Thank you for helping me

I was not commenting on your English. I was commenting on the fact that you think taking the square root of a negative number is a problem.

Since you now seem to have a difficulty in converting a number from the polar form $z = r cis (\theta)$ to rectangular form $z = x + iy$ I think my comment is well-founded.

Also, z=(cos 178,9 + isin 178,9) is not correct. z^2 = cis(120 degrees) or cis(-120 degrees). This leads to two values for z, neither of which is cis(178,9).

Sorry, but the evidence at the moment is that you do not have an adequate understanding of any of the background material that this question assumes that you know.

I suggest you review your class notes or textbook on how to do the conversion between polar and cartesion forms. I also suggest you review how to find the polar form of a complex number. You probably should also review how to complete the square.

**Plato** · March 12th 2011, 01:33 PM

Originally Posted by fgg

I wonder how to solve this complex math problem with algebra:
z^4+z^2+1=0

You might note that $z^4+z^2+1=(z^2+z+1)(z^2-z+1)$ .
Solve the two quadratics.

**fgg** · March 12th 2011, 01:40 PM

Originally Posted by mr fantastic

I was not commenting on your English. I was commenting on the fact that you think taking the square root of a negative number is a problem.

Since you now seem to have a difficulty in converting a number from the polar form $z = r cis (\theta)$ to rectangular form $z = x + iy$ I think my comment is well-founded.

Also, z=(cos 178,9 + isin 178,9) is not correct. z^2 = cis(120 degrees) or cis(-120 degrees). This leads to two values for z, neither of which is cis(178,9).

Sorry, but the evidence at the moment is that you do not have an adequate understanding of any of the background material that this question assumes that you know.

I suggest you review your class notes or textbook on how to do the conversion between polar and cartesion forms. I also suggest you review how to find the polar form of a complex number. You probably should also review how to complete the square.

I really regret comming to this forum. You are really insulting me. I don't think you know the answer to this and that is why you are so unpolite.

I surely know how to complete the square, I wrote 1 insted of 3/4 and in z=x+yi is -1/2 + square root of 3/ 2 i.

Anyway that is not the correct answer. Good bye

**mr fantastic** · March 12th 2011, 01:59 PM

Originally Posted by fgg

I really regret comming to this forum. You are really insulting me. I don't think you know the answer to this and that is why you are so unpolite.

I surely know how to complete the square, I wrote 1 insted of 3/4 and in z=x+yi is -1/2 + square root of 3/ 2 i.

Anyway that is not the correct answer. Good bye

1. I am not insulting you, I am stating the facts suggested by your posts. feel free to ignore my advice on reviewing the necessary background material.

2. Regarding completing the square, I am not a mind-reader. All I can do is comment on what gets posted. What you posted was wrong and suggested a lack of understanding.

3. z is NOT 1/2 + square root of 3/ 2 i. In fact z^2 is 1/2 + square root of 3/ 2 i and the other solution is z^2 is 1/2 - square root of 3/ 2 i. I have given you the polar form of both of these. The fact that you do not realise that there are TWO solutions for z^2 is grounds for further concern. The fact that the answers I gave you in polar form were for z^2 NOT z would explain why they are not the correct solutions for z!!

Knowing the two values for z^2 in polar form, you are expected to be able to use De Moivre's theorem to get the TWO values of z associated with each value of z^2. Once you have the four values of z in polar form, you are expected to be able to correctly convert them into exact rectangular form (using known exact values of sin and cosine). You will get the same values as the ones you get by solving the two quadratic equations suggested in Plato's latest reply (of course, if you cannot factorise the given equation into two irreducible factors, that particular method will not be very useful to you).

4. You are the one being insulting by saying that I don't know how to answer this question. There are many questions I don't know the answer to - yours is NOT one of them.

From what I have seen in this thread, you have a number of fundamental weaknesses. The fact is that you do not understand the material you are meant to know in order to correctly answer the question you posted. This is not my fault and it is not insulting to tell you this. I have suggested you go back and review that material. That is not insulting, in my view it is sensible advice. Whether or not you do so is your choice. But I honestly cannot see you being able to correctly answer this sort of question if you don't.

**HallsofIvy** · March 12th 2011, 02:31 PM

Originally Posted by fgg

Well my English may not be the best since I'm from Scandinavia, sorry about that. What I meant by not a "reell" number is that I can not take the square root of a negative number since z is a real number.

\
Again, you titled this "complex numbers". Why should z be a real number?

After using Moivres formula I will have
z=(cos 178,9 + isin 178,9) but if I have to answer in the form z=x + yi
How would you then proceed?

That is "in the form z= x+ yi". Assuming that "178,9" is in degrees, cos(178,9)=-0.9998 and sin(178,9)= 0.01920. Your answer is z= 0.9998+ 0.01920 i.

Thank you for helping me

**fgg** · March 12th 2011, 03:20 PM

Originally Posted by mr fantastic

1. I am not insulting you, I am stating the facts suggested by your posts. feel free to ignore my advice on reviewing the necessary background material.

2. Regarding completing the square, I am not a mind-reader. All I can do is comment on what gets posted. What you posted was wrong and suggested a lack of understanding.

3. z is NOT 1/2 + square root of 3/ 2 i. In fact z^2 is 1/2 + square root of 3/ 2 i and the other solution is z^2 is 1/2 - square root of 3/ 2 i. I have given you the polar form of both of these. The fact that you do not realise that there are TWO solutions for z^2 is grounds for further concern. The fact that the answers I gave you in polar form were for z^2 NOT z would explain why they are not the correct solutions for z!!

Knowing the two values for z^2 in polar form, you are expected to be able to use De Moivre's theorem to get the TWO values of z associated with each value of z^2. Once you have the four values of z in polar form, you are expected to be able to correctly convert them into exact rectangular form (using known exact values of sin and cosine). You will get the same values as the ones you get by solving the two quadratic equations suggested in Plato's latest reply (of course, if you cannot factorise the given equation into two irreducible factors, that particular method will not be very useful to you).

4. You are the one being insulting by saying that I don't know how to answer this question. There are many questions I don't know the answer to - yours is NOT one of them.

From what I have seen in this thread, you have a number of fundamental weaknesses. The fact is that you do not understand the material you are meant to know in order to correctly answer the question you posted. This is not my fault and it is not insulting to tell you this. I have suggested you go back and review that material. That is not insulting, in my view it is sensible advice. Whether or not you do so is your choice. But I honestly cannot see you being able to correctly answer this sort of question if you don't.

Thank you for answering despite my ugly tone. I really owe you an apology.Thank you so much!!! I see the solution now. Of course you were right, I had it written all wrong from the start z^2= -(1/2)+-(square root of 3 / 2)i

Yepp, you are my hero!

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