Discussion

Let \(z = 1 + i\), where \(i = \sqrt{-1}\). If \(z - \dfrac{24\bar{z}}{z^{2}} = \lambda z\), then the value of \(\lambda\) is equal to

Answer: B

\(\lambda = 13\).

With \(z = 1 + i\): \(\bar{z} = 1 - i\) and \(z^{2} = 1 + 2i + i^{2} = 2i\).

\[ \begin{aligned} \frac{24\bar{z}}{z^{2}} &= \frac{24(1 - i)}{2i} = \frac{12(1 - i)(-i)}{1} \\ &= 12(-i + i^{2}) = -12 - 12i \end{aligned} \]

\[ \begin{aligned} z - \frac{24\bar{z}}{z^{2}} &= 1 + i + 12 + 12i \\ &= 13 + 13i = 13(1 + i) \end{aligned} \]

So the expression equals \(13z\) and \(\lambda = 13\). The option 12 forgets the original \(z\) that is added back.

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