Do you recall synthetic or polynomial division? One way is to factor out from and you will get a linear equation.
Edit: you don't even need to go this complicated: just use the fact that is an even function, if z1 is a root, so is -z1.
This is not really a calculus question but I am taking a course in basic complex analysis and this is from the first chapter.
Show that the indicated numbers satisfy the given equation and find an additional solution, .
Obviously you can plug in and get a true statement, but how do you go about finding an additional solution?
Do you recall synthetic or polynomial division? One way is to factor out from and you will get a linear equation.
Edit: you don't even need to go this complicated: just use the fact that is an even function, if z1 is a root, so is -z1.
Why won't the conjugate of z1 work? It is indeed a zero when I do plugged it in. Perhaps there is some theory I'm not seeing here, all the op's question asks is to find an additional solution. And so z1 conjugate is that solution.
No it isn't. Recheck your calculations.
The theory your not seeing here is that roots won't magically always be conjugate pairs. I could just create an arbitrary polynomial which clearly violates that rule, for example, by multiplying out the following:
You are missing the fact that the coefficients of the polynomial have to be real in order for the roots to come in conjugate pairs.
The other root in this case is actually the negative of z1, not the conjugate.
We could probably just multiply z1 by e^(2kpi) for k in Z? Thanks sword for fixing my error, I didn't multiply correctly lol. Do you see an error with this new idea, using the fact that e^(2kpi)=1 is an identity.
Yes, having already determined one root, the simplest way to find the other is "division". If you wished to solve from the beginning, use polar form. This equation is the same as and -i has modulus 1 and argument (or ). To find the square root, take the square root of the modulus (1) and divide the argument by 2: so that one square root of -i is which is the root we have.
Since adding " " to the argument does not change the number, we can also write as modulus 1, argument . That gives, as a second root, .
In other words, the second root is just the negative of the first root, just as you learned way back in basic algebra!