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I have to solve these problems and show how to solve them in front of students tomorrow though I haven't learned the way to calculate infinite series of complex numbers.

(1) $\sum _{n=0} ^{\infty} z^n$

(2) $\sum _{n=0} ^{\infty} (3^{n+1} - 2^{n+2}) z^n$

Could you tell me the solutions or hints? Thank you in advance.

ynn
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    Both of these are geometric series (though the second one is a combination of two geometric series). These work the same in the complex case as in the real case. – Ian Oct 19 '16 at 13:49
  • @lan Thank you. Now I understand these problems aren't essentially different from those of real number. – ynn Oct 19 '16 at 14:18

1 Answers1

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Let $S_z=\sum_{n=0}^\infty z^n:$

$$(1-z)S_z=(1-z)(1+z+z^2+\dots)\\=(1\color{red}{+z+z^2+z^3+\dots})-(\color{red}{z+z^2+z^3+\dots})\\=1$$

$$S_z=\frac1{1-z}$$

Which is only true if $|z|<1$ so the sum converges.

$$\sum_{n=0}^\infty(3^{n+1}-2^{n+2})z^n=3\sum_{n=0}^\infty(3z)^n-4\sum_{n=0}^\infty(2z)^n$$

Which is of the form

$$=3S_{3z}-4S_{2z}$$

And then use the above with the condition $|z|<1/3$

Simply Beautiful Art
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  • Could you tell me the answer to (2)? I got the result of (6z-1)/((1-3z)(1-2z)) but WolframAlpha says the answer is z(6z+1)/((1-3z)(1-2z)). – ynn Oct 19 '16 at 14:20
  • I'm sorry but I made a mistake when I used WolframAlpha. It seems that my answer is correct. Thank you again. – ynn Oct 19 '16 at 14:22
  • @ynn :) You can hit the "answered my question" checkmark button below the voting buttons. It helps other users know if an answer is still open or not. – Simply Beautiful Art Oct 19 '16 at 17:04