Complex numbers: solving equation

Here's what I've done so far.

Using the second equation:

Substituting into the first equation:

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

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

Re: Complex numbers: solving equation

An alternate approach would be to rewrite the right side:

Re: Complex numbers: solving equation

To continue your method, write your quadratic in as:

Since *y* must be real, we are left with:

Re: Complex numbers: solving equation

Quote:

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 . Solve for and get suitable values for .

Re: Complex numbers: solving equation

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

is , so it's "really" . Solving (I'll just factor it):

, so or .

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: , and so forth.

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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 , where . Then

, where k is any integer.

Then , so or .

Note that , so

.

Thus or .

Thus or .

Thus or .

Thus or .

Triangle is , so .

Thus or .

Thus or .

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

Re: Complex numbers: solving equation

Thank you all for the answers.

Re: Complex numbers: solving equation