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Summation of a complex number from 0 to infinity
- Thread starter junam
- Start date
Yes I get it as 3/(3-2i), is this the solution.
There is no 'n' term in the formula, so I am little confused.
Say we have to find the the sum of (2i/3) to the power n frorm n=1 to n=3, when I tryo do it separately without using the formula
I get the answer (10i-12)/27.
So not sure how to go about it.
Thank you.
There is no 'n' term in the formula, so I am little confused.
Say we have to find the the sum of (2i/3) to the power n frorm n=1 to n=3, when I tryo do it separately without using the formula
I get the answer (10i-12)/27.
So not sure how to go about it.
Thank you.
Thank you very much Romsek, this is now resolved.
junam, find out if the problem/instructor requires you to write the final answer in \(\displaystyle \ \)a + bi form.
\(\displaystyle \dfrac{9 + 6i}{13} \ = \)
\(\displaystyle \dfrac{9}{13} + \dfrac{6}{13}i\)