imaginary numbers

$\displaystyle z(1+i) = 3 + 7i$ ,

so

$\displaystyle z = \frac{3+7i}{1+i}$ .

Multiply top and bottom by the conjugate of the denominator.

z=5+2i

z+iz= 3+7i How do you get z?

$\begin{array}{lrcl} & z+iz &=& 3+7i \\ & z(1+i) &=& 3+7i \\\\ \text{We set } & z = a+bi \\ & (a+bi)\cdot (1+i) &=& 3+7i \\ & a + ai+bi+bi^2 &=& 3+7i \qquad i^2 = -1\\ & a + ai+bi -b &=& 3+7i \\ & (a -b)+ (a+b) i &=& 3+7i \\\\ \text{We compare } &(1)\quad (a -b) &=& 3 \\ &(2)\quad (a+b) &=& 7 \\\\ & (a-b)+(a+b) &=& 3+7 \\ & a-b+a+b &=& 10 \\ & 2a &=& 10 \\ & \mathbf{a} & \mathbf{=} & \mathbf{5} \\\\ & (a+b)-(a-b) &=& 7-3 \\ & a+b-a+b &=& 4 \\ & 2b &=& 4 \\ & \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ & \mathbf{ z = a+bi }&\mathbf{=}& \mathbf{5+2i} \end{array}$