1. ## Complex numbers help

In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

PROBLEM 1

Find Re(w) and Im(w) [where $\displaystyle z=x+yi$]:

$\displaystyle w=z+\frac{1}{z}$

So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($\displaystyle x$ and $\displaystyle y$).

My way would be:

$\displaystyle w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{ 2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=$
$\displaystyle \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$

If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.

However, the final solution is:

$\displaystyle Re(w)=x+\frac{x}{x^{2}+y^{2}}$ and $\displaystyle Im(w)=y-\frac{y}{x^{2}+y^{2}}$

Does anybody know the steps leading to the final solution?

PROBLEM 2

Find all solutions to the equation [where $\displaystyle z=x+yi$]:

$\displaystyle 2z^{2}-3\overline{z}^{2}=10i$

It is a quadratic equation and has a conjugate of complex number in it.

2. Originally Posted by summerBoy
In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

PROBLEM 1

Find Re(w) and Im(w) [where $\displaystyle z=x+yi$]:

$\displaystyle w=z+\frac{1}{z}$

So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($\displaystyle x$ and $\displaystyle y$).

My way would be:

$\displaystyle w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{ 2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=$
$\displaystyle \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$

If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.
.
You did everything correct except expand which is important.

For example if you have,
(1+i)(2+i)/(25)
Then if you expand,
(1+3i)/25=(1/25)+(3/35)i
Thus,
Re(z)=1/25 and Im(z)=3/35

Note, this is the the solution to your problem but this is the general way to do it.

3. Originally Posted by summerBoy
In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

PROBLEM 1

Find Re(w) and Im(w) [where $\displaystyle z=x+yi$]:

$\displaystyle w=z+\frac{1}{z}$

So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ($\displaystyle x$ and $\displaystyle y$).

My way would be:

$\displaystyle w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{ 2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=$
$\displaystyle \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...$

If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.

However, the final solution is:

$\displaystyle Re(w)=x+\frac{x}{x^{2}+y^{2}}$ and $\displaystyle Im(w)=y-\frac{y}{x^{2}+y^{2}}$

Does anybody know the steps leading to the final solution?

PROBLEM 2

Find all solutions to the equation [where $\displaystyle z=x+yi$]:

$\displaystyle 2z^{2}-3\overline{z}^{2}=10i$

It is a quadratic equation and has a conjugate of complex number in it.

Oh, it can be opened now? I tried to answer this since many hours ago but it wouldn't open.

Anyway, I tried the 1st Problem only. I did expand it paiently and I came up to the shown Final solution.

Here goes:

W = [((x +yi)^2 +1)(x -yi)] / (x^2 +y^2)

W = [(x^2 +2xyi -y^2 +1)(x -yi)] / (x^2 +y^2)

W = [x^3 -(x^2)yi +2(x^2)yi -2x(y^2)(i^2) -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

W = [x^3 -(x^2)yi +2(x^2)yi +2xy^2 -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

W = [x^3 +(x^2)yi +xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

Separate the imaginaries, or collect the reals and the imaginaries separately
W = [(x^3 +xy^2 +x) +i(yx^2 +y^3 -y)] / (x^2 +y^2)

So,
Re(W) = (x^3 +xy^2 +x) /(x^2 +y^2)
Im(W) = [(y^3 +yx^2 -y) /(y^2 +x^2)](i).

Do long divisions on both and you'd arrive to the shown solution.

4. Thanks for help.

5. Originally Posted by summerBoy
Find all solutions to the equation [where $\displaystyle z=x+yi$]:

$\displaystyle 2z^{2}-3\overline{z}^{2}=10i$

It is a quadratic equation and has a conjugate of complex number in it.

Important note: x and y are assumed to be real numbers!!!

We have:
2(x + iy)^2 - 3(x - iy)^2 = 10i

2x^2 + 4ixy - 2y^2 - 3x^2 + 6ixy + 3y^2 = 10i

(-x^2 + y^2 ) + i(10xy - 10) = 0

Now, both the real and imaginary parts of this expression must be 0 so we have the system of equations:
-x^2 + y^2 = 0
10xy - 10 = 0

I would solve the bottom equation for y:
10xy = 10
xy = 1
y = 1/x

and insert this into the top equation:
-x^2 + (1/x)^2 = 0
-x^4 + 1 = 0
x^4 = 1

Now, remember that x is a REAL number. There are only two real solutions to this equation:
x = -1, 1

So y = 1/x means that y = -1, 1 respectively.

Thus your solution set is (-1, -1), (1, 1). Thus z = -1 - i and z = 1 + i. You can check these yourself.

-Dan

6. Thank you very much for your help.