Finding argument of complex number

[songoku]

Homework Statement

For $z+\frac 1 z = 1+ 2i$, find $|z^2 - \frac{1}{z^2}|$

Relevant Equations

conjugate

modulus of complex number: $|z|=\sqrt{x^2+y^2}$

Let: $z=x+iy$

$$z+\frac 1 z =1+2i$$
$$x+iy +\frac{1}{x+iy}=1+2i$$
$$x+iy+\frac{1}{x+iy} . \frac{x-iy}{x-iy}=1+2i$$
$$x+iy+\frac{x-iy}{x^2+y^2}=1+2i$$
$$\frac{x^3+xy^2+x+i(x^2y+y^3-y)}{x^2+y^2}=1+2i$$

So:
$$\frac{x^3+xy^2+x}{x^2+y^2}=1$$
$$x^3+xy^2+x=x^2+y^2$$

and
$$\frac{x^2y+y^3-y}{x^2+y^2}=2$$
$$x^2y+y^3-y=2(x^2+y^2)$$

Combining the two equations: $x^2y+y^3-y=2(x^3+xy^2+x)$

How to solve that equation? Or maybe there is another approach to this question?

Thanks

 

[haruspex]

Homework Statement:: For $z+\frac 1 z = 1+ 2i$

Not sure it's the best, but an obvious thing to try is to solve that quadratic

 

[DaveE]

Not sure it's the best, but an obvious thing to try is to solve that quadratic

Yes, I think dealing with $z$ instead of $x + iy$ is best. Also note that $z^2- \frac{1}{z^2} = (z+\frac{1}{z})(z-\frac{1}{z})$ may help. That's how I did it.

It does seem like brute force to me though. I bet someone has a more clever solution.

 

[haruspex]

It does seem like brute force to me though. I bet someone has a more clever solution.

Yes, I suspect there's a neat way which avoids finding z.
I tried $|z^2-z^{-2}|=\frac{|z^2-1|}{|z|}\frac{|z^2+1|}{|z|}=\frac{|z^2-1|}{|z|}|1+2i|$, but it wasn't obvious where to go from there.

 

[songoku]

Not sure it's the best, but an obvious thing to try is to solve that quadratic

Yes, I think dealing with $z$ instead of $x + iy$ is best. Also note that $z^2- \frac{1}{z^2} = (z+\frac{1}{z})(z-\frac{1}{z})$ may help. That's how I did it.

It does seem like brute force to me though. I bet someone has a more clever solution.

I will try this and brute force it. Thank you very much haruspex and DaveE

 

[DaveE]

There is a relatively simple solution. I show it to you after you make an attempt yourself with these hints:

1) Stay with z, not the rectangular or polar forms.
2) I preferred to say $Z_o \equiv z+\frac{1}{z} = 1+2i$ but this was mostly to save writing $1+2i$ repeatedly.
3) Solve the quadratic equation to get roots $Z_{1,2}$ in terms of $Z_o$.
4) Factor $z^2-\frac{1}{z^2}$ as in my previous post. Simplify this function by substituting $Z_o$ as much as possible.
5) Substitute the quadratic roots $Z_{1,2}$ into the simplified form of $z^2-\frac{1}{z^2}$.

You will get a relatively simple answer without ever actually putting in the value of $Z_o$ until the end. The algebra should never get really messy.

If you get stuck, describe where/what/how and I'll help.

 

[Sunay_]

I think this solution is fairly easy
$(z+\frac{1}{z})^2 = -3+4i$
Then $z^2+2+ \frac{1}{z^2} = -3+4i$
if you subtract 4 from both sides it is $(z-\frac{1}{z})^2 = -7 + 4i$
Now we know what $(z^2-\frac{1}{z^2})^2$ is
and from here it can easily be calculated.

 

[songoku]

if you subtract 4 from both sides it is $(z-\frac{1}{z})^2 = -7 + 4i$
Now we know what $(z^2-\frac{1}{z^2})^2$ is...

Do you mean: "now we know what $(z-\frac{1}{z})^2$ is" ?

There is a relatively simple solution. I show it to you after you make an attempt yourself with these hints:

1) Stay with z, not the rectangular or polar forms.
2) I preferred to say $Z_o \equiv z+\frac{1}{z} = 1+2i$ but this was mostly to save writing $1+2i$ repeatedly.
3) Solve the quadratic equation to get roots $Z_{1,2}$ in terms of $Z_o$.
4) Factor $z^2-\frac{1}{z^2}$ as in my previous post. Simplify this function by substituting $Z_o$ as much as possible.
5) Substitute the quadratic roots $Z_{1,2}$ into the simplified form of $z^2-\frac{1}{z^2}$.

You will get a relatively simple answer without ever actually putting in the value of $Z_o$ until the end. The algebra should never get really messy.

If you get stuck, describe where/what/how and I'll help.

I tried this and I think I also got the same result using method suggested by @Sunay_

In the end, I need to find $\sqrt{-7+4i}$ and then multiply by $1+2i$ to get $z^2-\frac{1}{z^2}$ and after that finding the modulus.

I also tried brute force and it almost cost me my life

Using @DaveE method:
$$z+\frac 1 z=Z_o$$
$$z^2-Z_o z+1=0$$
$$z=\frac{Z_o \pm \sqrt{Z_o^2-4}}{2}$$

Then
$$z^2-\frac{1}{z^2}=\left(z+\frac 1 z \right)\left(z - \frac 1 z\right)$$
$$=Z_o(2z-Z_o)$$
$$=Z_o(2 . \frac{Z_o \pm \sqrt{Z_o^2-4}}{2} - Z_o)$$
$$=\pm Z_o \sqrt{Z_o^2 - 4}$$

Edit: Wait, I think I understand what @Sunay_ meant by we know $\left(z^2-\frac{1}{z^2}\right)^2$. I will try and compare

 

[DaveE]

Do you mean: "now we know what $(z-\frac{1}{z})^2$ is" ?I tried this and I think I also got the same result using method suggested by @Sunay_

In the end, I need to find $\sqrt{-7+4i}$ and then multiply by $1+2i$ to get $z^2-\frac{1}{z^2}$ and after that finding the modulus.

I also tried brute force and it almost cost me my life

Using @DaveE method:
$$z+\frac 1 z=Z_o$$
$$z^2-Z_o z+1=0$$
$$z=\frac{Z_o \pm \sqrt{Z_o^2-4}}{2}$$

Then
$$z^2-\frac{1}{z^2}=\left(z+\frac 1 z \right)\left(z - \frac 1 z\right)$$
$$=Z_o(2z-Z_o)$$
$$=Z_o(2 . \frac{Z_o \pm \sqrt{Z_o^2-4}}{2} - Z_o)$$
$$=Z_o \sqrt{Z_o^2 - 4}$$

Yes, although you left out the $\pm$ part at the end, because of the square root. Of course it doesn't matter if you only need the magnitude.

Both methods are basically the same. When @Sunay_ "completed the square" that was equivalent to the quadratic formula. I think her method is a little nicer, maybe shorter, but I wouldn't have a good argument why.

 

[songoku]

Yes, although you left out the $\pm$ part at the end, because of the square root. Of course it doesn't matter if you only need the magnitude.

Yes, I just realized it

Both methods are basically the same. When @Sunay_ "completed the square" that was equivalent to the quadratic formula. I think her method is a little nicer, maybe shorter, but I wouldn't have a good argument why.

I also think @Sunay_ method is shorter but not everytime I can think of that method by myself so I am okay with longer method (as long as it is easier for me to find it) although it takes more time during the test

Thank you very much for all the help and explanation haruspex, DaveE and Sunay_

 

[PeroK]

Here's another variation:
$$z + \frac 1 z = w \ \Rightarrow \ 2z = w \pm \sqrt{w^2 - 4} \ \ (quadratic)$$
$$2z = (z + \frac 1 z) + (z - \frac 1 z) = w + (z - \frac 1 z)$$
$$\therefore \ z - \frac 1 z = \pm \sqrt{w^2 - 4}$$
$$z^2 - \frac 1 {z^2} = (z + \frac 1 z)(z - \frac 1 z) = \pm w\sqrt{w^2 - 4}$$