First slide

Geometry of complex numbers

Question

$It is given that complex numbers z_{1} and z_{2} satisfy |z_{1}| = 2 and |z_{2}| = 3 .If the included angle$ $of their corresponding vectors is 60^{\circ} then |\frac{z_{1} + z_{2}}{z_{1} - z_{2}}| can be expressed as \frac{\sqrt{N}}{7}, where N is a$ $natural number, then N =$

Moderate

Solution

$Given that |z_{1}| = 2 and |z_{2}| = 3$
By using cosine rule
$\begin{array}{l} |z_{1} + z_{2}| = \sqrt{{|z_{1}|}^{2} + {|z_{2}|}^{2} + 2 |z_{1}| |z_{2}| \cos 60^{0}} \\ = \sqrt{4 + 9 + 2 (3)} = \sqrt{4 + 9 + 6} = \sqrt{19} \end{array}$
$\begin{array}{l} and |z_{1} - z_{2}| = \sqrt{{|z_{1}|}^{2} + {|z_{2}|}^{2} - 2 |z_{1}| |z_{2}| \cdot \cos 60^{\circ}} \\ = \sqrt{4 + 9 - 6} = \sqrt{7} \end{array}$
$\begin{array}{l} |\frac{z_{1} + z_{2}}{z_{1} - z_{2}}| = \sqrt{\frac{19}{7}} = \sqrt{\frac{19 \times 7}{7^{2}}} = \frac{\sqrt{133}}{7} = \frac{\sqrt{N}}{7} (given) \\ ∴ N = 133 \end{array}$

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