# real part series and imaginary part series

Theorem 1. Given the series

 $\displaystyle z_{1}+z_{2}+z_{3}+\ldots$ (1)

with the real parts of its terms  $\Re{z_{n}}=a_{n}$  and the imaginary parts of its terms  $\Im{z_{n}}=b_{n}$  ($n=1,\,2,\,3,\,\ldots$). If the series (1) converges and its sum is $A+iB$, where $A$ and $B$ are real, then also the series

 $a_{1}+a_{2}+a_{3}+\ldots\;\;\;\mbox{and}\;\;\;b_{1}+b_{2}+b_{3}+\ldots$

converge and their sums are $A$ and $B$, respectively. The converse is valid as well.

Proof. Let $\varepsilon$ be an arbitrary positive number. Denote the partial sum of (1) by

 $S_{n}=z_{1}+\ldots+z_{n}=(a_{1}+ib_{1})+\ldots+(a_{n}+ib_{n})=(a_{1}+\ldots+a_% {n})+i(b_{1}+\ldots+b_{n}):=A_{n}+iB_{n}$

($n=1,\,2,\,3,\,\ldots$). When (1) converges to the sum $A+iB$, then there is a number $n_{\varepsilon}$ such that  for any integer  $n>n_{\varepsilon}$  we have

 $|(A_{n}-A)+i(B_{n}-B)|=|(A_{n}+iB_{n})-(A+iB)|<\varepsilon.$

But a complex number is always absolutely at least equal to the real part (see the inequalities in modulus of complex number), and therefore  $|A_{n}-A|\leqq|(A_{n}-A)+i(B_{n}-B)|<\varepsilon$, similarly  $|B_{n}-B|\leqq|(A_{n}-A)+i(B_{n}-B)|<\varepsilon$  as soon as  $n>n_{\varepsilon}$.  Hence,  $A_{n}\to A$  and  $B_{n}\to B$  as  $n\to\infty$.  This means the convergences

 $a_{1}+a_{2}+a_{3}+\ldots=A\;\;\;\mbox{and}\;\;\;b_{1}+b_{2}+b_{3}+\ldots=B,$

Q.E.D. The converse part is straightforward.

Theorem 2. Notations same as in the preceding theorem. The series

 $|z_{1}|+|z_{2}|+|z_{3}|+\ldots$

converges if and only if the series

 $a_{1}+a_{2}+a_{3}+\ldots\;\;\;\mbox{and}\;\;\;b_{1}+b_{2}+b_{3}+\ldots$

converge absolutely (http://planetmath.org/AbsoluteConvergence).

Proof. Use the inequalities

 $0\leqq|a_{n}|\leqq|z_{n}|,\quad 0\leqq|b_{n}|\leqq|z_{n}|$

and

 $0\leqq|z_{n}|\leqq|a_{n}|+|b_{n}|$

for using the comparison test.

Theorem 3. If the series $\displaystyle\sum_{n=1}^{\infty}|z_{n}|$ converges, then also the series $\displaystyle\sum_{n=1}^{\infty}z_{n}$ converges and we have

 $\left|\sum_{n=1}^{\infty}z_{n}\right|\leqq\sum_{n=1}^{\infty}|z_{n}|.$

Proof. By theorem 2, the convergence of $\sum|z_{n}|$ implies the convergence of $\sum a_{n}$ and $\sum b_{n}$, which, by theorem 1, in turn imply the convergence of $\sum z_{n}$ . Since for every $n$ the triangle inequality guarantees the inequality

 $\left|\sum_{j=1}^{n}z_{j}\right|\leqq\sum_{j=1}^{n}|z_{j}|,$

then we must have the asserted limit inequality, too.

Title real part series and imaginary part series RealPartSeriesAndImaginaryPartSeries 2013-03-22 17:28:08 2013-03-22 17:28:08 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 40-00 SumOfSeries ModulusOfComplexNumber AbsoluteConvergenceTheorem RealAndImaginaryPartsOfContourIntegral