Complex Quadratic?

Hey guys, I'm a bit unsure on this question.





I tried to write out the equation with everything in modulus-argument form and manage to find the r (modulus) values however when I try and equate the arguments I'm not sure if I'm doing it right. I think it's the fact that there are both z^2 and z terms involved which is throwing me off. Here's what I've done so far.

$iz^{2}+2z=i-1$

Mod-Arg form:

$i[r(cos{\theta}+isin{\theta})]^{2}+2[r(cos{\theta}+isin{\theta})]=\sqrt{2}(cos{\frac{3\pi}{4}}+isin{\frac{3\pi}{4}})$

Using De Moivre's Theorem:

$ir^{2}(cos{2\theta}+isin{2\theta})+2r(cos{\theta}+isin{\theta})=\sqrt{2}(cos{\frac{3\pi}{4}}+isin{\frac{3\pi}{4}})$

Comparing the Moduli:

$ir^{2}+2r=\sqrt{2}$

Equating Real Parts:

$r^{2}=0$
$r=0$

Equating Imaginary Parts:

$2r=\sqrt{2}$
$r=\frac{\pm\sqrt{2}}{2}$

So we have the two possible moduli values. For the finding the argument values, is it as simple as equating the thetas? So:

$2\theta+\theta=\frac{3\pi}{4}$
$3\theta=\theta=\frac{3\pi}{4}$
$\theta=\frac{9\pi}{4}, \frac{13\pi}{4}$

Is this right? If so, I know how to do the rest by simply converting my two mod-arg numbers into cartesian form.

I think I'm making this harder than it actually is but I'm still coming to different answers using both of your methods. Finishing mine:

$z=\frac{\sqrt2}{2}cos{\frac{9\pi}{4}}+\frac{\sqrt2}{2}sin{\frac{9\pi}{4}}i$

and

$z=\frac{\sqrt2}{2}cos{\frac{13\pi}{4}}+\frac{\sqrt2}{2}sin{\frac{13\pi}{4}}i$

so

$z_1=\frac{1}{2}+\frac{1}{2}i$
$z_2=-\frac{1}{2}-\frac{1}{2}i$

I'm not getting the same answers using the quadratic formula or by subbing expanding out z terms. Can someone confirm that my answer is correct?

No, those are not correct. Post your working?

Oh right, all of the working is now in the OP.

I literally just used my theta values of

$\theta=\frac{9\pi}{4}, \frac{13\pi}{4}$

and my r value of

$r=\frac{\sqrt{2}}{2}$

to get the two complex roots in mod-arg form. I then converted them to cartesian form to get the final roots.

Then you must have an error earlier in your working.

My working..

$\displaystyle i z^2+2z-i+1=0$

I multiplied through by -i. Not really necessary but..

$\displaystyle z^2-2iz-1-i=0$

Then use the formula:

$\displaystyle z=\frac{2i\pm\sqrt{-4-4(-1-i)}}{2}$

$\displaystyle z=i\pm\sqrt{i}$

$\displaystyle z=i\pm\frac{\sqrt{2}}{2}(1+i)$

etc.

How have you gone from your second last to last steps?

Modulus and argument.