Discussion
If \(z + \bar{z} = 6\) and \(z - \bar{z} = 4i\) , then \(|z|^{2}\)=
Answer: D
\(|z|^{2} = \) 13.
With \(z = x + iy\), \(z + \bar{z} = 2x\) and \(z - \bar{z} = 2iy\). So \(2x = 6\) gives \(x = 3\), and \(2iy = 4i\) gives \(y = 2\).
\[ |z|^{2} = x^{2} + y^{2} = 9 + 4 = 13 \]
36 and 16 square the sums 6 and 4 instead of the parts; 9 is \(x^{2}\) alone; 15 does not arise from any of the parts.
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