+0

# iMagiNarY NumBeRs

0
297
2
+25

Find the complex number z that satisfies

$$(1 + i) z - 2 \overline{z} = -11 + 25i.$$

May 18, 2021

#1
0

z = 5 - 2i.

May 19, 2021
#2
+26322
+3

Find the complex number z that satisfies
$$(1 + i) z - 2 \overline{z} = -11 + 25i$$.

$$\text{Set z=a+bi} \\ \text{Set \overline{z}=a-bi}$$

$$\begin{array}{|rcll|} \hline \mathbf{ (1 + i) z - 2 \overline{z} } &=& \mathbf{ -11 + 25i } \\ (1 + i)(a+bi) - 2(a-bi) &=& -11 + 25i \\ (1 + i)a+(1+i)bi - 2a+2bi &=& -11 + 25i \\ a + ia+bi+bi^2 - 2a+2bi &=& -11 + 25i \quad | \quad i^2 = -1 \\ a + ia+bi-b - 2a+2bi &=& -11 + 25i \\ -(a+b) +i(a+3b)&=& -11 + 25i \qquad \text{compare}\\ \hline -(a+b) &=& -11 \\ a+b &=& 11 \\ \mathbf{a} &=& \mathbf{11-b} \\ \hline (a+3b) &=& 25 \\ 11-b+3b &=& 25 \\ 2b &=& 14 \\ \mathbf{b} &=& \mathbf{7} \\ \hline \mathbf{a} &=& \mathbf{11-b} \\ a &=& 11-7 \\ \mathbf{a} &=& \mathbf{4} \\ \hline z &=& a+bi \\ \mathbf{z} &=& \mathbf{4+7i} \\ \hline \end{array}$$

May 19, 2021