Complex numbers: solving equation

**theintervurt** · September 11th 2012, 10:15 PM

Here's what I've done so far.

$\displaystyle \begin{align*} z^2 = -8 + 6i \\(x + yi)^2 = -8 + 6i \\ (x^2 - y^2) + (2xy)i = -8 + 6i \\ \\ x^2 - y^2 = -8 \textrm{ and } 2xy = 6 \end{align*}$

Using the second equation:

$\displaystyle \begin{align*} x = \frac{3}{y} \end{align*}$

Substituting into the first equation:

$\displaystyle \begin{align*} \frac{9}{y^2} - y^2 = -8 \\ 9 - y^4 = -8y^2 \\ -y^4 + 8y^2 + 9 = 0 \end{align*}$

And that is where I'm stuck, I thought of using a horner scheme but still didn't make it

P.S: Could someone indicate how to correctly set the alignment of the last equation ?

**MarkFL** · September 11th 2012, 10:23 PM

An alternate approach would be to rewrite the right side:

$-8+6i=1+2\cdot3i+9i^2=(1+3i)^2$

**MarkFL** · September 11th 2012, 10:33 PM

To continue your method, write your quadratic in $y^2$ as:

$y^4-8y^2-9=0$

$\left(y^2-9 \right)\left(y^2+1 \right)=0$

Since y must be real, we are left with:

$(y+3)(y-3)=0$

**kalyanram** · September 11th 2012, 10:34 PM

Originally Posted by theintervurt

-y^4 + 8y^2 + 9 = 0 \end{align*} [/tex]

You can continue this line of thought thought. This is a quadratic in $y^2$ . Solve for $y^2$ and get suitable values for $x$ .

**johnsomeone** · September 11th 2012, 10:52 PM

You're fine - you just need to know a useful little trick to continue. Some 3-term polynomials are actually quadratics in disguise.

$y^4 - 8y^2 - 9 = 0$ is $(y^2)^2 - 8(y^2)^1 - 9 = 0$ , so it's "really" $u^2 - 8u - 9 = 0$ . Solving (I'll just factor it):

$(y^2 - 9)(y^2 + 1) = 0$ , so $y^2 = 9$ or $y^2 = -1$ .

When you start with z = x+iy, you're saying x and y are real numbers. Thus you can ignore the imaginary solutions for y, leaving only y = 3 and y = -3.

This trick can be useful. Similar examples: $q^{14} + 5q^7 + 6 = 0, 4w^{100} - 21w^{50} + 20 = 0, r^{8/3} + 5r^{4/3} - 5 = 0$ , and so forth.

-----

By the way, your approach is completely fine, but can also do it by putting everything into polar form (this approach works better with higher powers than quadratics):

There's an ERROR somewhere in what follows, but the approach is correct:

Let $z=Re^{i\theta}, -8+6i = 10e^{i\alpha}$ , where $\alpha = arctan(6/(-8)) = arctan(-.75)$ . Then

$z^2 = -8+6i \implies R^2e^{i2\theta} = 10e^{i(\alpha + 2\pi k)} \implies R = \sqrt{10}, 2\theta = \alpha + 2k\pi$ , where k is any integer.

Then $\theta = \alpha/2 + \pi k$ , so $z = \sqrt{10}e^{i\alpha/2}$ or $z = \sqrt{10}e^{i ( \alpha /2+ \pi )}$ .

Note that $tan(\alpha) = -.75$ , so $tan(\alpha /2) = \frac{tan(\alpha)}{1+\sqrt{1+tan^2(\alpha)}} = \frac{(-3/4)}{1+\sqrt{1+(-3/4)^2}}$

$= \frac{(-3/4)}{1+\sqrt{25/16}} = \frac{(-3/4)}{1+(5/4)} = \frac{(-3/4)}{(9/4)} = -1/3$ .

Thus $z = \sqrt{10}e^{i arctan(-1/3)}$ or $z = \sqrt{10}e^{i ( arctan(-1/3) + \pi )}$ .

Thus $z = \sqrt{10}e^{- i arctan(1/3)}$ or $z = -\sqrt{10}e^{-i arctan(1/3)}$ .

Thus $z = \sqrt{10}(cos(-arctan(1/3)) + i sin( -arctan(1/3))$ or $z = -\sqrt{10}(cos(-arctan(1/3)) + i sin( -arctan(1/3))$ .

Thus $z = \sqrt{10}(cos(arctan(1/3)) - i sin(arctan(1/3))$ or $z = \sqrt{10}(-cos(arctan(1/3)) + i sin(arctan(1/3))$ .

Triangle is $1, 3, \sqrt{10}$ , so $cos(arctan(1/3)) = 3/\sqrt{10}, sin(arctan(1/3)) = 1/\sqrt{10}$ .

Thus $z = \sqrt{10}((3/\sqrt{10}) - i (1/\sqrt{10})$ or $z = \sqrt{10}( -(3/\sqrt{10})+ i (1/\sqrt{10})$ .

Thus $z = 3 - i$ or $z = -3 + i$ .

Ugg - I blundered in here somewhere. It should be 1 + 3i and -1 - 3i. But the approach is correct, and good to know.

**theintervurt** · September 11th 2012, 11:52 PM

Thank you all for the answers.

**MaxJasper** · September 12th 2012, 12:20 AM

$z=\pm \sqrt{-8+6i}=\pm (1+3i)$

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