Solve Complex Number Equation: z^2+4z ̅+4=0 | Expert Help Available

[ronho1234]

find all the solutions to z^2+4z ̅+4=0 where z is a complex number.

 

[Mentallic]

Is the quadratic $z^2+4z+4$? Simply use the quadratic formula.

 

[ronho1234]

i think I'm getting confused so is the answer for z=2 or do i have to represent it as complex 2isquared? because the question asks for all solution so i don't think 2 is right... and what does the bar on top of the z mean?

 

[Mentallic]

i think I'm getting confused so is the answer for z=2 or do i have to represent it as complex 2isquared? because the question asks for all solution so i don't think 2 is right... and what does the bar on top of the z mean?

Oh I see, so you were trying to solve $z^2+4\overline{z}+4=0$ then? Because in your OP I'm just seeing a box that is being replaced by whatever you wanted to write down.

If it's just $z^2+4z+4$ then you can simply factorize it into $(z+2)^2=0$ which would give you the answer of z=-2, but this can't be right because the question already stated that z is a complex number (imaginary part is not 0 by assumption).

If it's $z^2+4\overline{z}+4=0$ then you'll need to use the fact that $z=a+ib$ and $\overline{z}=a-ib$ and plug these values into the expression and then simplify and equate the real and imaginary parts.
What I mean by equating them is if you end up with something like $ab+2+(a-1)i=0$ then what you can take away from this is that $ab+2=0$ and $a-1=0$ because the LHS is equal to the RHS thus the real part on the LHS is equal to 0 on the RHS, and the imaginary part on the LHS is equal to the imaginary part on the RHS (the RHS is equal to 0+0i).

 

[ronho1234]

so i partially understand what you have
so i did something like this:
z = a+ib and zbar=a-ib
because i split them like you said (z+2i)(z-2i)=0
so i made a+ib=z+2 and a-ib=zbar+2
so z=a+ib-2 and zbar=a-ib-2
i put these back into th eqn (a+ib-2)squared +4(a-ib-2)+4=0
then i expand and simplify i get a squared+2aib-8ib-b-4=0
and so (a squared -b-4)+2i(ab-4b)=0?
then a squared -b=4 and ab=4b
so a=4 and b=12?
then so i get z=4+12i and zbar=4-12i ?
i don't get why we have to split it as z and zbar...

 

[Mentallic]

Whew that was a tough one to follow.

so i did something like this:
z = a+ib and zbar=a-ib
because i split them like you said (z+2i)(z-2i)=0

Ok stop right there. Where did I say to "split them" as such? If $z=a+ib$ and $\overline{z}=a-ib$ then substituting these values into

$$z^2+4\overline{z}+4=0$$

gives us

$$(a+ib)^2+4(a-ib)+4=0$$

Now expand, simplify, and collect the real and imaginary parts and equate them.

 

[ronho1234]

i reread your explanation and redid it
so what i did was sub a+ib and a-ib straight into z^2+4zbar+4=0
and i got a squared -b squared +4 + 4a +2bi(a-2) = o
and then a-2=0 a=2 and a squared -b squared +4+4a=0
and i get a=2 and b=+/-4
do i sub these back into z and z bar or something?

 

[ronho1234]

so my solutions are: z=2+4i and z=2-4i??

 

[Curious3141]

so my solutions are: z=2+4i and z=2-4i??

And z=-2. Real numbers are a subset of the complex plane.

 

[Curious3141]

i reread your explanation and redid it
so what i did was sub a+ib and a-ib straight into z^2+4zbar+4=0
and i got a squared -b squared +4 + 4a +2bi(a-2) = o

So far so good.

and then a-2=0 a=2

OR b=0.

That's also a valid solution.

and a squared -b squared +4+4a=0
and i get a=2 and b=+/-4
do i sub these back into z and z bar or something?

When you sub b=0 back into this, you get a² + 4a + 4=0, which will give you a=-2, b=0; the only real solution z=-2 which I mentioned in my last post.

So there are three solutions in total.

 

[ronho1234]

yep... so my solutions are z=2+4i, z=2-4i and z=-2
thankyou!

 

[NewtonianAlch]

In the future use punctuation and be neater so people can help you easily!