Complex numbers: find the values of and

Question

If $z=x+jy$, where $x$ and $y$ are real, find the values of $x$ and $y$ when

$\frac{3z}{1-j}+\frac{3z}{j}= \frac{4}{3-j}.$

My attempt:

$\frac{3(x+jy)}{1-j}+\frac{3(x+jy)}{j}= \frac{4}{3-j}$

$\frac{3x+3jy}{1-j}+\frac{3x+3jy}{j}= \frac{4}{3-j}.$

I'm not sure how to go on after this step above. Please help; I have an exam in a few hours.

Answer 1

$$\frac {3z}{1-j}+\frac{3z}j=\frac4{3-j}$$

$$\frac {3zj+3z(1-j)}{(1-j)j}=\frac4{3-j}$$

$$\frac{3z}{1+j}=\frac4{3-j}$$

$$3z=\frac{4(1+j)}{3-j}=\frac{4+4j}{3-j}=\frac{(4+4j)(3+j)}{10}=\frac{(2+2j)(3+j)}5=\frac{4+8j}5$$

$$z=\dfrac{4+8j}{15}$$

Answer 2

Since\begin{align}\frac{3z}{1-i}+\frac{3z}i&=3z\left(\frac1{1-i}+\frac1i\right)\\&=3z\left(\frac12-\frac i2\right)\\&=z\left(\frac32-\frac32i\right),\end{align}you have\begin{align}z&=\frac{\frac4{3-i}}{\frac32-\frac32i}\\&=\frac{\frac{6}{5}+\frac{2 i}{5}}{\frac32-\frac32i}\\&=\frac4{15}+\frac8{15}i.\end{align}

Answer 3

The complex equation is $$\frac{3x+3jy)}{1-j}+\frac{3x+3jy}{j}=\frac{4}{3-j}.$$ Here $j^2=-1$. Multiply first, second and third term term by $1+j,-j$, and $3+j$ up and down and separate real and imaginary parts as $$(-6/5+3x/2+3y/2)+j(-2/5-3x/2+3y/2)=0,\implies x+y=4/5,-x+y=4/15$$ these two simultaneous Eqs. then give $x=4/15,y=8/15$.