# Solving complex equation

How do i further solve the following complex equation:

$$z\cdot \bar{z} + z + \bar{z} + i\cdot z - \overline{i \cdot z} = 9 + 4i$$ $$a^{2} - b^{2} + 2a - 2b = 9 + 4i$$

How do i solve from here on ?

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Do you mean $z\bar{z}+z+\bar{z}+iz-\overline{iz}=9+4i$? –  André Nicolas Aug 29 '12 at 14:29
Yes my latex is bad i didnt know how to do that negate sign. –  Sterling Duchess Aug 29 '12 at 14:30

To solve the equation $$z\bar{z}+z+\bar{z}+iz-\overline{iz}=9+4i,$$ we can proceed much as you did. Let $z=a+ib$.

Then $z\bar{z}=(a+ib)(a-ib)=a^2+b^2$.

We have $iz=-b+ia$, so its conjugate is $-b-ia$. Our expression is therefore equal to $$a^2+b^2+(a+ib)+(a-ib)+(-b+ia)-(-b-ia),$$ which simplifies to $a^2+b^2+2a+2ia$.

This is $9+4i$ precisely if the imaginary parts match and the real parts match. We end up with the equations $2a=4$ and $a^2+b^2+2a=9$. Now $a$ and then $b$ are easy to find.

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To finish the job, a = 2 and b = 1 :) –  Martin Aug 29 '12 at 14:44
@Martin: Well, there is also $b=-1$. –  André Nicolas Aug 29 '12 at 14:46
So $$2ai = 4i, a^{2} + b^{2} + 2a = 9$$ Would then a = 2i and i put this a in to $$a^2{2} + b^{2} + 2a = 9$$ to solve for b ? –  Sterling Duchess Aug 29 '12 at 14:48
@kellax: I wrote $2a=4$. If you like, you can write $2ai=4i$ and divide both sides by $i$, getting $2a=4$. Then $a=2$. Now because $a^2+b^2+2a=9$, we have $2^2+b^2+4=9$, so $b^2=1$, and therefore $b=\pm 1$. –  André Nicolas Aug 29 '12 at 14:52
Thank you very much. –  Sterling Duchess Aug 29 '12 at 14:56

Alternative approach: the real part of both hands must be equal: $$z\bar z + z+\bar z = 9.$$ The imaginary part of both hands must be equal: $$i(z+\bar z)=4i.$$

So $z$ and $\bar z$ are two numbers; their sum is $4$ and their product is $9-4=5$. So they satisfy $$z^2 - 4z + 5 = 0.$$ This you can solve.

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