The complex numbers a and b satisfy a \(\overline{b} = -1 + 5i.\) Find \(\overline{a} b.\)
Set \(a=c+di\) and \(b=r+si\) where \(c\), \(d\), \(r\), and \(s\) are real numbers. This makes the \(a\overline{b}=-1+5i\) turn into \((c+di)(r-si)=-1+5i\). When we multiply the left side out, we get \((cr+ds)+(dr-cs)i\). So, \(cr+ds=-1\) and \(dr-cs=5\).
Notice that \(\overline{a}b=(c-di)(r+si)=(cr+ds)+(cs-dr)i=(cr+ds)-(dr-cs)i\). Plugging in the values before gives us \(\overline{a}b=-1-5i\).
Hope this helps!
The complex numbers \(a\) and \(b\) satisfy \(a\overline{b} = -1 + 5i\).
Find \(\overline{a} b\).
\(\begin{array}{|rcll|} \hline a\overline{b} &=& -1 + 5i \\\\ \overline{a\overline{b}} &=& \overline{-1 + 5i} \quad | \quad \overline{\overline{b}} = b \\\\ \overline{a} b &=& \overline{-1 + 5i} \quad | \quad \overline{-1 + 5i} = -1-5i \\\\ \mathbf{\overline{a} b} &\mathbf{=}& \mathbf{-1-5i} \\ \hline \end{array}\)