Finding the solutions to a quadratic equation with a complex conjugate

[Cottontails]

Homework Statement

Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

Homework Equations

Alternate form: 4conjugate[z] + z^2 = -4

The Attempt at a Solution

I have tried solving this solution using the quadratic formula.
However, √b^2 - 4ac = √16 - 4x1x4 = 0. Therefore, as the square root is not negative, there are no imaginary numbers and the solution cannot be complex, right? Although, I am also confused with solving this, given that there is a conjugate in the equation. So, would I have to solve the equation twice, one as z^2+4z+4=0 and the other as z^2-4z+4=0?
I also put the equation into wolfram alpha and got the real solution as z=-2 and the complex solutions as z=2-4i, z=2+4i. Is that the right answer? How would you get the real solution and the complex solution from the equation then?

 

[tiny-tim]

Hi Cottontails!

(try using the X² button just above the Reply box )

z^2 + 4conjugate[z] + 4 = 0

hint: the complex conjugate of that equation (indeed, any true equation) is also true

 

[Cottontails]

Sorry, I don't understand. What do you mean by saying that the complex conjugate is true?

 

[Mark44]

Homework Statement

Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

Homework Equations

Alternate form: 4conjugate[z] + z^2 = -4

The Attempt at a Solution

I have tried solving this solution using the quadratic formula.
However, √b^2 - 4ac = √16 - 4x1x4 = 0. Therefore, as the square root is not negative, there are no imaginary numbers and the solution cannot be complex, right? Although, I am also confused with solving this, given that there is a conjugate in the equation. So, would I have to solve the equation twice, one as z^2+4z+4=0 and the other as z^2-4z+4=0?
I also put the equation into wolfram alpha and got the real solution as z=-2 and the complex solutions as z=2-4i, z=2+4i. Is that the right answer? How would you get the real solution and the complex solution from the equation then?

Strictly speaking, your equation is not quadratic in z (because of the presence of [itex]\overline{z}[/itex]. It would be a quadratic if it were z² + 4z + 4 = 0.

The way to go here is to write z = a + bi so [itex]\overline{z}[/itex] = a - bi. Substitute in the original equation and separate into real and imaginary parts, and then solve for a and b. This works to produce the same answers that wolframalpha provided.

 

[tiny-tim]

Sorry, I don't understand. What do you mean by saying that the complex conjugate is true?

[itex]\bar{z}^2 + 4z + 4 = 0[/itex]

(for the same z)

 

[Cottontails]

Strictly speaking, your equation is not quadratic in z (because of the presence of [itex]\overline{z}[/itex]. It would be a quadratic if it were z² + 4z + 4 = 0.

The way to go here is to write z = a + bi so [itex]\overline{z}[/itex] = a - bi. Substitute in the original equation and separate into real and imaginary parts, and then solve for a and b. This works to produce the same answers that wolframalpha provided.

Yes, I have tried that method as well however, was unable to finish solving it so I assumed I was solving it incorrectly.
z = a + bi conjugate[z]= a - bi
(a+bi)^2 + 4(a-bi) + 4 = 0
-> a^2 - bi^2 - 2abi + 4 =0
That is all I got up to. I do not know how to solve the equation from there.

 

[Mark44]

Yes, I have tried that method as well however, was unable to finish solving it so I assumed I was solving it incorrectly.
z = a + bi conjugate[z]= a - bi
(a+bi)^2 + 4(a-bi) + 4 = 0
-> a^2 - bi^2 - 2abi + 4 =0

You have a couple of mistakes above, possibly due to skipping steps.
For one mistake, (bi)² = -b², not -bi².

Start by carefully expanding this equation: (a+bi)² + 4(a-bi) + 4 = 0

That is all I got up to. I do not know how to solve the equation from there.

 

[Cottontails]

You have a couple of mistakes above, possibly due to skipping steps.
For one mistake, (bi)² = -b², not -bi².

Start by carefully expanding this equation: (a+bi)² + 4(a-bi) + 4 = 0

Yes, sorry. I missed adding in the expansion of 4(a-bi). I also thought about bi^2 = -b^2 (as i^2=-1) yet, got confused as to whether it should be left or not. Thanks for the correction there.
Okay, so now I have:
a^2 + 2abi - b^2 + 4a - 4bi + 4 = 0
So, from here, would I have to factorise to reduce the equation further? As, I can't find any common factors.

 

[Mark44]

Yes, sorry. I missed adding in the expansion of 4(a-bi). I also thought about bi^2 = -b^2 (as i^2=-1) yet, got confused as to whether it should be left or not. Thanks for the correction there.
Okay, so now I have:
a^2 + 2abi - b^2 + 4a - 4bi + 4 = 0
So, from here, would I have to factorise to reduce the equation further? As, I can't find any common factors.

No, split this equation into its real part and its imaginary part, which will result in two equations. At that point you will need to do some easy factoring. Then solve for a and b.

 

[Cottontails]

No, split this equation into its real part and its imaginary part, which will result in two equations. At that point you will need to do some easy factoring. Then solve for a and b.

Okay, thanks. I understand that you solve for a and b and those values will the be substituted into z = a + bi and z = a - bi to get the two complex solutions.
However, how you find the real solution (in finding z = -2)? I obtained a = -2 from solving the imaginary parts. So, would you then use b = 0 and then substitute that into z = (-2) + (0)i to then give z = -2?
As for solving the real parts, I have gotten stuck with it.
So, the equation: a^2 - b^2 + 4a + 4 = 0
a^2 = b^2 - 4a - 4
a = √b^2 - 4a - 4
a = b - 2√a - 2
b^2 = a^2 + 4a + 4
b = √a^2 + 4a + 4
b = a + 2√a + 2
Then substituting b into a:
a = (a + 2√a + 2) - 2√a - 2
Yet, that then equals a = a
With doing the same for substituting a into b, I also get b = b.
I know I'm doing something wrong here so, what is my mistake?
(I previously solved the equation, mistakingly using the equation a^2 - b^2 + 4 =0 and I was able to understand how it was able to give the values of a and b (yet, I know that's not right as the equation is wrong). So, I do understand the method required but just can't recognise where I went wrong with my working.

 

[Mark44]

Okay, thanks. I understand that you solve for a and b and those values will the be substituted into z = a + bi and z = a - bi to get the two complex solutions.
However, how you find the real solution (in finding z = -2)? I obtained a = -2 from solving the imaginary parts. So, would you then use b = 0 and then substitute that into z = (-2) + (0)i to then give z = -2?
As for solving the real parts, I have gotten stuck with it.
So, the equation: a^2 - b^2 + 4a + 4 = 0

Write this as a² + 4a + 4 - b² = 0.
or (a + 2)² - b² = 0.
Can you continue from there?

a^2 = b^2 - 4a - 4
a = √b^2 - 4a - 4
a = b - 2√a - 2
b^2 = a^2 + 4a + 4
b = √a^2 + 4a + 4
b = a + 2√a + 2
Then substituting b into a:
a = (a + 2√a + 2) - 2√a - 2
Yet, that then equals a = a
With doing the same for substituting a into b, I also get b = b.
I know I'm doing something wrong here so, what is my mistake?
(I previously solved the equation, mistakingly using the equation a^2 - b^2 + 4 =0 and I was able to understand how it was able to give the values of a and b (yet, I know that's not right as the equation is wrong). So, I do understand the method required but just can't recognise where I went wrong with my working.

 

[Cottontails]

Write this as a² + 4a + 4 - b² = 0.
or (a + 2)² - b² = 0.
Can you continue from there?

(a + 2)^2 - b^2 = 0
b^2 = (a + 2)^2
b = a + 2
However, how would I solve it to get the value of a? I have tried substituting b = a + 2 back into the equation but then I get 0, which I know isn't right...

 

[tiny-tim]

Hi Cottontails!

Slow dowwwwn, and go back to …

Okay, so now I have:
a^2 + 2abi - b^2 + 4a - 4bi + 4 = 0

(try using the X² button just above the Reply box )

a² + 2abi - b² + 4a - 4bi + 4 = 0

is two simultaneous equations …

so (general strategy ) solve the easy one first, then substitute that into the other one

in this case, the easy one is 2abi - 4bi = 0 ​

 

[Cottontails]

Hi Cottontails!

Slow dowwwwn, and go back to …


(try using the X² button just above the Reply box )

a² + 2abi - b² + 4a - 4bi + 4 = 0

is two simultaneous equations …

so (general strategy ) solve the easy one first, then substitute that into the other one

in this case, the easy one is 2abi - 4bi = 0 ​

Okay, thanks for pointing that out. I solved 2abi - 4bi = 0
2abi = 4bi
-> a = 2
b = a + 2
b = 2 + 2
-> b = 4
z = a + bi, z = a - bi
z = 2 + 4i, z = 2 - 4i are the imaginary/complex solutions.
However, how would you find the real solution (z = -2)? Would you just use the original equation and solve it using the quadratic formula?

 

[tiny-tim]

Okay, thanks for pointing that out. I solved 2abi - 4bi = 0
2abi = 4bi
-> a = 2

aha! wrooong!

the correct statement is …

2abi = 4bi

-> a = 2 or b = 0​

… innit?

(the moral: never divide by 0 and expect the universe not to notice )

 

[Cottontails]

Yes, that makes sense. However, if a = 2 and b = 0, then if you substitute it back into z = a + bi, then you get z = 2, which doesn't satisfy the imaginary or real solutions. Yet, if you substitute b = 0 into b = a + 2, you get a = -2. Substituting that into z = a + bi, it would then equal z = -2 (real solution).
Sorry, I'm really confused now...

 

[tiny-tim]

-> a = 2 or b = 0

… However, if a = 2 and b = 0 …

"or" … "and" …

get some sleep! :zzz:​

 

[Cottontails]

Oh, right. So, from solving the imaginary parts, you get b = 0. You substitute that into b = a + 2, to get a = -2. Therefore, substituting that into z = a + bi, z = -2 (real solution).
For the imaginary solution, it is how I had solved it previously.
Is that now all correct?

 

[tiny-tim]

yup!

 

[Cottontails]

Okay, great. Thank you for all your help!

 

[Curious3141]

Just wanted to say that this is the exact same problem posted in another recent thread (on the same pageview) where Mentallic and I helped: https://www.physicsforums.com/showthread.php?t=598945

You guys in the same class or something?

 

[Mark44]

(a + 2)^2 - b^2 = 0
b^2 = (a + 2)^2
b = a + 2

You missed a solution, and I don't think any of the other people responding noticed.

If you factor the first equation above, you get
((a + 2) - b)((a + 2) + b) = 0

This leads to a + 2 = ± b

However, how would I solve it to get the value of a? I have tried substituting b = a + 2 back into the equation but then I get 0, which I know isn't right...

 

[Cottontails]

You missed a solution, and I don't think any of the other people responding noticed.

If you factor the first equation above, you get
((a + 2) - b)((a + 2) + b) = 0

This leads to a + 2 = ± b

Oh, wow. Thanks for pointing that out!

 

[Mark44]

The mistake you made is thinking that the equation x² = 4 can be solved by taking the square root of each side, and getting x = 2.

A better way to approach this is to write x² - 4 = 0, or (x - 2)(x + 2) = 0, which results in x = 2 or x = -2.