I have the equation: $z^2 +2az -1 =0$, where $z$ is a complex variable and $a$ is a complex parameter. To find the solutions of $z$, I just used the quadratic equation, and obtained:

$z_1 = -a + (1+a^2)^{(1/2)}$

$z_2 = -a - (1+a^2)^{(1/2)}$

This is homework (I'm sorry if that's frowned upon), and the ultimate goal is to find the radius of convergence of $z_1$ and $z_2$ after converting the roots to a taylor series. I only know of ways to do this if there is a nice sigma notation form of the series. I can calculate the first few terms:

$z_1$: $1 + x + x^2/2 - x^4/8 + x^6/16 - 5x^8/128$

The second root is similar in that it lacks an apparent pattern. I'm unsure of how I'm supposed to go about finding a radius of convergence without sigma notation of how the series 'works'. Have I approached the problem wrong?


  • $\begingroup$ Put superscript arguments of more than 1 character in brackets {}. $\endgroup$ – user_194421 Jan 30 '18 at 4:39

You don't really need an explicit form to determine radius of convergence. You have $z_1=-1+(1+a^2)^{1/2}$. When $u$ is not a nonnegative integer, the binomial series $$(1+x)^u=\sum_{n=0}^\infty\frac{u(u-1)(u-2)\cdots(u-n+1)}{n!}x^n$$ has radius of convergence $1$; therefore the same is true for the Maclaurin series of $(1+x^2)^{1/2}$, and so the series for $z_1$ also has radius of convergence $1$.


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