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I don't have experience with solving these types of equations.

After looking around; I learned that I should try substituting a + bi in for a complex number in these types of equations.

$$w^2 + 3w^* + 2 = 0$$

I found that the real part is $a^2 - b^2 + 3a + 2$

I also found the imaginary part to be $2ab - 3b$

I'm not sure how to proceed.

Note, w* is the complex conjugate.

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So, equating the imaginary parts,

$$b(2a-3)=0$$

Equating the real parts, $$0=a^2-b^2+3a+2$$ Can you take it from here?

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  • $\begingroup$ I believe so. You solve for a and b using the "imaginary" equation as it's easier. You use the two values found and put them in the "real equation"; this generates another four values. You then have a and b for the four a + bi solutions. $\endgroup$ – Timeless Cookie Mar 17 at 15:52
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Just for fun, here's another approach:

$$w^2+3w^*+2=0\implies \begin{cases} w^*=-(w^2+2)/3\quad\text{(by solving for }w^*)\\ (w^*)^2+3w+2=0\quad\text{(by conjugating)} \end{cases}$$

and thus

$${(w^2+2)^2\over9}+3w+2=0$$

or, after expanding, simplifying, and factoring,

$$w^4+4w^2+27w+22=(w+1)(w+2)(w^2-3w+11)=0$$

for which the roots are

$$w=-1,-2,\text{ and }{3\pm\sqrt{-35}\over2}$$

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  • $\begingroup$ I'll review conjugation. It's the only part I don't know in your solution; basically where w and w* swap. $\endgroup$ – Timeless Cookie Mar 17 at 16:06
  • $\begingroup$ @TimelessCookie, the main thing to learn is that $(w^*)^*=w$. That and the fact that conjugation "distributes" over algebraic operations is all you need to know. It comes in useful from time to time. $\endgroup$ – Barry Cipra Mar 17 at 16:24
  • $\begingroup$ @TimelessCookie You could take a look at this video (4:30 onwards) and the second half where the basic properties of complex conjugates are reviewed/proven. $\endgroup$ – Ryan G Mar 17 at 16:58

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