# Finding the complex square by solving

1. Apr 13, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

i) Find the complex square roots of -24-10$\iota$ by solving (x+$\iota$y)$^{2}$ = -24 -10$\iota$ for x,y real.

ii) Hence, find the roots of the quadratic $\zeta$$^{2}$ -(1+$\iota$)$\zeta$ +(6+3$\iota$) in "a+$\iota$b" form where a,b are real numbers

By the way does anyone know where I can find the correct latex symbols for z & i?? Sorry for the ones used here but I thought they were the correct one's to use until I saw the post.

2. Relevant equations

3. The attempt at a solution

So I started by expanding the square, and seperating to real and imag parts..

x$^{2}$-y$^{2}$=-24 $\rightarrow$ 2x$\iota$y =-10$\iota$

and x$^{2}$+y$^{2}$= $\sqrt{(24^{2}+10^{2}}$ = 26

solving for x$^{2}$=-24+y$^{2}$

then subing into eqn x$^{2}$+y$^{2}$

therefore y$^{2}$=25 $\rightarrow$ y=5

so x$^{2}$=1 since x$^{2}$+25=26

so roots are $\pm$(1+5$\iota$)

I think this should be $\pm$(1-5$\iota$) but dont understand why could someone please explain as I always get confused here.

So for part ii)

I used quadratic formula for the eqn

$\zeta$$^{2}$ -(1+$\iota$)$\zeta$ +(6+3$\iota$)

So by simplifying I get

$\frac{1+\iota\pm\sqrt{-24+10\iota}}{2}$

Now replacing the sqrt term with my term from part i)

I now have $\frac{1+\iota\pm(1+5\iota)}{2}$

So suming up terms I have my two roots of..

1+3$\iota$ & -2$\iota$

Is this right I dont think so because my understanding is that I could sub these terms back into the quadratic eqn and I should get 0... but I dont??

2. Apr 13, 2012

### Mentallic

You can just type in the entire equation into latex. You don't need to keep switching between latex and normal text for everything that you want:

So for the quadratic you posted, we would have

$$z^2-(1+i)z+6+3i$$

Is written out by typing [ tex ]z^2-(1+i)z+6+3i[ /tex ] (without the spaces in the tex tags)

But these tags extend to the next line, so you can use the itex tags like you've been doing if you want them in-line with everything else.

Let's first start with the two equations that you had from expanding $(x+iy)^2$

x$^{2}$-y$^{2}$=-24 $\rightarrow$ 2x$\iota$y =-10$\iota$

Notice in the second equation, it can be simplified into $$x=\frac{-5}{y}$$ and what this means is that if x is positive, then y must be negative, and vice versa. They are opposite signs of each other.
You can't extract this information from squares however because squaring a value loses its sign. This is why I would advise against using the modulus $x^2+y^2=r^2$ as a way of solving this kind of problem. What you can instead do is:

$$x^2-y^2=-24$$

$$2xyi=-10i$$

Simplifying and making either variable the subject:

$$y=\frac{-5}{x}$$

Plugging this value back into the first equation:

$$x^2-\left(\frac{-5}{x}\right)^2=-24$$

Simplifying:

$$x^4-25=-24x^2$$

$$x^4+24x^2-25=0$$

This is a quadratic in x2 that has rational roots, so you can either make the substitution z=x2 to make things more clear as to how it's a quadratic, but at this level you should be able to skip that substitution entirely, so after factorizing we have:

$$(x^2+25)(x^2-1)=0$$

And since we began with the assumption that x is real, the only valid factor is $x^2-1=0 \rightarrow x=\pm 1$

Now what you want to do is substitute this back into that earlier equation that didn't have squares, $$y=\frac{-5}{x}$$

So what we get is $y\mp 5$

Thus the solutions are $\pm1\mp 5i$ or equivalently, $\pm(1-5i)$

Now, the reason you didn't get the correct answer is because of the line where you said
This is not true. If $y^2=25$ then $y=\pm 5$, and we would need to do extra work to determine whether y=5, -5 or possibly both are acceptable solutions. It turns out that y=-5 is the only one that works.

Are you sure you get this? You should be getting -24-10i for the discriminant.

And thus your final answer is wrong because $\sqrt{-24-10i}=\pm(1-5i)$ as we showed in the previous question.

3. Apr 13, 2012

### charmedbeauty

ok sorry I do get $-24 -10i$

But when I find my result I should be able to check it by plugging into the quadratic and should get 0!?

by the way that is a much better way i finding x & y! thanks for the help!

4. Apr 14, 2012

### Mentallic

Yes, of course! That's what solving the quadratic means! We're looking to find the values of z such that $$z^2-(1+i)z+6+3i=0$$ (there will be two values).

5. Apr 15, 2012

### charmedbeauty

thanks!

Last edited: Apr 15, 2012