complex number problem

**hyzlemon** · March 16th 2010, 05:25 AM

solve for x and y:
x + yi = 2/3+i + 3/2+i

**apcalculus** · March 16th 2010, 06:42 AM

Originally Posted by hyzlemon

solve for x and y:
x + yi = 2/3+i + 3/2+i

Combine the imaginary and real parts on the right hand side:

x + y i = (2/3 + 3/2) + 2 i

y = 2

and

x = ...

Good luck!

**Plato** · March 16th 2010, 07:22 AM

Originally Posted by hyzlemon

solve for x and y:
x + yi = 2/3+i + 3/2+i

Is the problem $x+yi=\frac{2}{3+i}+\frac{3}{2+i}?$

**Soroban** · March 16th 2010, 08:45 AM

Hello, hyzlemon!

Solve for $x$ and $y$ : . $x + yi \:=\: \frac{2}{3+i} + \frac{3}{2+i}$

Rationalize the right side:

. . $\frac{2}{3+i}\cdot{\color{blue}\frac{3-i}{3-i}} + \frac{3}{2+i}\cdot{\color{blue}\frac{2-i}{2-i}} \;=\;\frac{2(3-i)}{9+1} + \frac{3(2-i)}{4+1} \;=\;\frac{6-2i}{10} + \frac{6-3i}{5}$

. . . $=\;\frac{6-2i}{10} + \frac{12-6i}{10} \;=\;\frac{18-8i}{10} \;=\;\frac{9}{5} - \frac{4}{5}i$

Therefore: . $x + yi \;=\;\frac{9}{5} - \frac{4}{5}i \quad\Rightarrow\quad \begin{Bmatrix}x &=& \dfrac{9}{5} \\ \\[-3mm] y &=& \text{-}\dfrac{4}{5} \end{Bmatrix}$

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