Solving Complex Equations: Find x+yi & arg(z+2i)=π/4

[pinsky]

Homework Statement

Find all the complex numbers defined with:

\left | z^2 - 1 \right | = \left | z \right |^2
arg(z+2i) = \pi / 4

Homework Equations

z=x + yi

agr(z)=ArcTg(y/x)

The Attempt at a Solution

I've changed z with x+yi but it turned out to be a forth order equation to get solutions for y. So I'm guessing that isn't the right way.

Any hints?

 

[epenguin]

For the first of these if you got a 4th deg equation it looks like you have forgotten about the square root that appears in the definition of modulus.

You must not forget either the two-valued nature of a square root!

You don't even need to "change z with x+yi".

It will also help you make sense if you think about these problems geometrically, even if you do not do your calculations that way. Either before or after the calculation think of the simple geometry.

 

[pinsky]

Damn, I've falsely written the first equation.
It should go

<br /> \left | z^2 - 1 \right | = \left | z \right |^2 -4<br />

Sorrym, i haven't really understood your tip.
Could you repeat that more clearly please?

 

[epenguin]

Write the definition of |anything|.

From that write out the equations without the || that you can get from

<br /> \left | z^2 - 1 \right | = \left | z \right |^2 -4<br />

Without changing the z into anything else.

 

[pinsky]

I'm not following again.

\left |z| = \sqrt{z\bar{z}}

Is that the definition of a module without writing z as x+yi?

I supose i can remove the module from |z|². But what do i get then?

<br /> <br /> \left | z^2 - 1 \right | = z^2 -4<br /> <br />

 

[Char. Limit]

I'm not following again.

\left |z| = \sqrt{z\bar{z}}

Is that the definition of a module without writing z as x+yi?

I supose i can remove the module from |z|². But what do i get then?

<br /> <br /> \left | z^2 - 1 \right | = z^2 -4<br /> <br />

|z| = sqrt(x^2 + y^2)

use this to rewrite your equations in terms of x and y.

 

[pinsky]

If i take the solution in that way, i get a forth order equation. That isn't something that is solvable in the time the exam lasts.

 

[Char. Limit]

If i take the solution in that way, i get a forth order equation. That isn't something that is solvable in the time the exam lasts.

Well I don't see another easy way to do it, other than guessing... and guessing often doesn't work.

 

[pinsky]

For the first of these if you got a 4th deg equation it looks like you have forgotten about the square root that appears in the definition of modulus.

You must not forget either the two-valued nature of a square root!

You don't even need to "change z with x+yi".

Do you understand what epenguin suggests?

 

[epenguin]

I supose i can remove the module from |z|². But what do i get then?

<br /> <br /> \left | z^2 - 1 \right | = z^2 -4<br /> <br />

Now carefully remove the module from the other side!

(I frankly don't know how it occurs to you to do it on one side and doesn't on the other, but still.)

If the result doesn't make sense to you, see my first post.