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Don't let the $i$'s get to you! For instance, in 1(a), $\displaystyle\sum_{j=0}^{\infty} \left(\frac{i}{3}\right)^j$ is geometric since $\left|\dfrac{i}{3}\right| = \dfrac{|i|}{3} = \dfrac{1}{3} < 1$.Please refer to the attached image.
How do I do these questions, In particular
1a, 1c and 1f.
Could anyone give me a hint to get me started with either of these?
Thanks.
Note thatThanks for that.
Although, for
1a) it says that the sum of the series is
$\frac{1}{1-\frac{i}{3}}$ $ = $ $\frac{9+3i}{10}$
Why is this? I am unsure how they get to the first term of $\frac{1}{1-\frac{i}{3}}$
this i understood,Note that
\[\frac{1}{1-\dfrac{i}{3}} = \frac{3}{3\left(1-\dfrac{i}{3}\right)} = \frac{3}{3-i}\]
Now conjugate and you'll get the desired result.
thanks!Use that power series
\(\displaystyle \frac{1}{1-z} = \sum_{k=0}^{\infty}z^k \,\,\,\,\,\,\, |z|<1\)