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 Jun 1 '14 at 19:25
  • 1
    $\begingroup$ Hint: $\overline{(1+i)}=1-i$ and $\overline{(ab)}=\overline{a}\cdot\overline{b}.$ $\endgroup$ – M. Strochyk Jun 1 '14 at 19:26
  • $\begingroup$ I know the feeling. Complex numbers are both fascinating and complicated to understand. $\endgroup$ – Robert Soupe Jun 1 '14 at 19:26
  • $\begingroup$ Maybe write $c_1 = x+iy$ and solve for $x,y$? $\endgroup$ – copper.hat Jun 1 '14 at 19:27

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 $c_1 = {1 \over 10}+i{11 \over 10}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.