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Find the complex number z that satisfies

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

 May 18, 2021
 #1
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z = 5 - 2i.

 May 19, 2021
 #2
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Find the complex number z that satisfies
\((1 + i) z - 2 \overline{z} = -11 + 25i\).

 

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

 

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

 

laugh

 May 19, 2021

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