This might seem quite trivial for people who are knowledgeable in complex analysis, but it is not so much to me.

I am trying to find an efficient way to solve the following system of equations: $$ \frac{11}{10} = c_1(1+i)^{-2} + \overline{c}_1(1-i)^{-2} \\ \frac{1}{2} = c_1(1+i)^{-3} + \overline{c}_1(1-i)^{-3} $$

The unknowns are obviously $c_1$ and its conjugate $\overline{c}_1$. I have tried multiplying the first equality by $(1+i)^{-1}$, substract it from the second one and so on, but it gets quite messy. I was hoping somebody could tell me a better approach.

Thanks very much

  • 1
    $\begingroup$ Note that $(1+i)^{-2}=-\frac{i}{2}$ and $(1-i)^{-2}=\frac{i}{2}$. Similarly $(1+i)^{-3}=-\frac{1+i}{4}$ and $(1-i)^{-3}=\frac{1-i}{4}$. $\endgroup$
    – Servaes
    Commented Jun 1, 2014 at 19:25
  • 1
    $\begingroup$ Hint: $\overline{(1+i)}=1-i$ and $\overline{(ab)}=\overline{a}\cdot\overline{b}.$ $\endgroup$ Commented Jun 1, 2014 at 19:26
  • $\begingroup$ I know the feeling. Complex numbers are both fascinating and complicated to understand. $\endgroup$ Commented Jun 1, 2014 at 19:26
  • $\begingroup$ Maybe write $c_1 = x+iy$ and solve for $x,y$? $\endgroup$
    – copper.hat
    Commented Jun 1, 2014 at 19:27

3 Answers 3


Let's write

$$z = \frac{c_1}{(1+i)^2}.$$

Then we have the equations

$$\frac{11}{10} = z + \overline{z} = 2\operatorname{Re} z$$


$$\frac{1}{2} = \frac{z}{1+i} + \overline{\left(\frac{z}{1+i}\right)} = 2\operatorname{Re} \frac{z}{1+i}.$$

Now note that

$$\frac{z}{1+i} = \frac{z(1-i)}{2}$$

to rewrite the second equation to

$$\frac{1}{2} = \operatorname{Re} \bigl( z(1-i)\bigr) = \operatorname{Re} z + \operatorname{Im} z.$$

From that it is not hard to determine $z$, and thence $c_1 = z(1+i)^2 = 2i z$.


Maybe it's less messy if you set $z=c_1$ and $w=1+i$. Then your equations are $$ \begin{cases} w^{-2}z+\bar{w}^{-2}\bar{z}=a\\ w^{-3}z+\bar{w}^{-3}\bar{z}=b \end{cases} $$ where $a=11/10$ and $b=1/2$. The idea of multiplying the first equation by $\bar{w}^{-1}$ and subtracting is good, because you get $$ w^{-2}(\bar{w}^{-1}-w^{-1})z=a\bar{w}^{-1}-b $$ so $$ z=w^2\frac{a\bar{w}^{-1}-b}{\bar{w}^{-1}-w^{-1}}= w^2\frac{a-b\bar{w}}{\bar{w}}\frac{w\bar{w}}{w-\bar{w}}= w^3\frac{a-b\bar{w}}{w-\bar{w}} $$ This is just computation.


Here is another approach:

Let $c_1 = x+iy$, and substitute into the equations above. After a little tedious manipulation, this gives ${11 \over 10} = y$ and ${1 \over 2} = {y-x \over 2}$.

We can read off the solution as $c_1 = {1 \over 10}+i{11 \over 10}$.


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