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Solve for $z$ in the following equation: $2-iz = 1 + 3iz$. 

[i]Express your answer in standard form[/i].

 Feb 21, 2021
 #1
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Solve for \(z\) in the following equation: \(2-iz = 1 + 3iz\).

 

\(\begin{array}{|rcll|} \hline 2-iz &=& 1 + 3iz \\ 2-1-iz-3iz &=& 0 \\ 1-4iz &=& 0 \quad | \quad \times i\\ 1*i-4i^2z &=& 0 \\ i-4i^2z &=& 0 \quad | \quad \mathbf{i^2 = -1} \\ i-4(-1)z &=& 0 \\ i+4z &=& 0 \\ 4z &=& -i \\ \mathbf{z} &=& \mathbf{-\dfrac{i}{4}} \\ \hline \end{array}\)

 

laugh

 Feb 21, 2021

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