Ytukyg

$dfrac{1}{z+3i}$ can be interpreted as the sum of the geometric series $displaystyledfrac{1}{6i}sum_{k=0}^{infty}left(frac{z-3i}{6i}right)^n$ this can be obtained by writing:
$dfrac{1}{z+3i}=dfrac{1}{6i+(z-3i)}=dfrac{1}{6i}dfrac{1}{1+dfrac{1}{6i}(z-3i)}$ now the book I'm reading says that this is equal to the series I said at the beginning. But how is it possible if the sum of a geometric series is in the form of $dfrac{1}{1-q}$?

Edit: I am sure there is convergence of the geometric sum because $|z-3i|<5$

Probably I don't fully understand your question. But let's make a try.

Consider the following series:

$$S = q + qx + qx^2 + qx^3 + ldots+qx^{n-1} $$

Now let's multiply the whole series by $x$:

$$Sx = qx + qx^2 + qx^3 + qx^4 + ldots+qx^n$$

Now let's subtract the second from the first:

$$S - Sx = (q + qx + qx^2 + qx^3 + ldots+qx^{n-1}) - (qx + qx^2 + qx^3 + qx^4 + ldots+qx^n)$$

We can easily show that only few terms survive at the right side, that is,

$$S(1 - x) = q - qx^n$$

Whence

$$S(1-x) = q(1-x^n)$$

$$S = qfrac{1-x^n}{1-x}$$

As $n$ goes to infinity, the absolute value of $x$ must be less than $1$ for the series to converge, the sum then becomes

$$q + qx + qx^2 + qx^3 + ldots = sum_{k = 0}^{+infty} qx^k = frac{q}{1-x} ~~~~~~~ |x| < 1$$

And for $q = 1$ we just get

$$frac{1}{1-x}$$

As wanted.