+1678 Solve for $z$ in the following equation: $2-iz = 1 + 3iz$.
[i]Express your answer in standard form[/i].
+26393 Solve for \(z\) in the following equation: \(2-iz = 1 + 3iz\).
\(\begin{array}{|rcll|} \hline 2-iz &=& 1 + 3iz \\ 2-1-iz-3iz &=& 0 \\ 1-4iz &=& 0 \quad | \quad \times i\\ 1*i-4i^2z &=& 0 \\ i-4i^2z &=& 0 \quad | \quad \mathbf{i^2 = -1} \\ i-4(-1)z &=& 0 \\ i+4z &=& 0 \\ 4z &=& -i \\ \mathbf{z} &=& \mathbf{-\dfrac{i}{4}} \\ \hline \end{array}\)
