Let $$\alpha = 8 - 14i,A = \left\{ {z \in c:{{\alpha z - \overline \alpha \overline z } \over {{z^2} - {{\left( {\overline z } \right)}^2} - 112i}}=1} \right\}$$ and $B = \left\{ {z \in c:\left| {z + 3i} \right| = 4} \right\}$. Then $$\sum\limits_{z \in A \cap B} {({\mathop{\rm Re}\nolimits} z - {\mathop{\rm Im}\nolimits} z)} $$ is equal to ____________.
Answer (integer)
14
Solution
<p>Let $z = x + iy$</p>
<p>and $\alpha = 8 - 14i$</p>
<p>$${{\alpha z - \overline \alpha \,\overline z } \over {{z^2} - {{\overline z }^2} - 112i}} = 1$$</p>
<p>$\therefore$ ${{(16y - 28x)} \over {4xy - 112i}} = 1$</p>
<p>$(16y - 28x + 112)i = 4xy$</p>
<p>$\therefore$ $z = - 7i$ or 4</p>
<p>Now, $z = - 7i$ satisfy B</p>
<p>$B:{x^2} + {(y + 3)^2} = 16$</p>
<p>$A \cap B = (0, - 7)$</p>
<p>${\mathop{\rm Re}\nolimits} z - lm\,z = 7$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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