Question:
If $ \frac{5{{z}_{2}}}{11{{z}_{1}}} $ is purely imaginary, then the value of $ \left[ \frac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right] $ is
If $ \frac{5{{z}_{2}}}{11{{z}_{1}}} $ is purely imaginary, then the value of $ \left[ \frac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right] $ is
Updated On: Jun 6, 2022
- $ \frac{37}{33} $
- $ 2 $
- 1
- $ 3 $
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The Correct Option is C
Solution and Explanation
Since, $ \frac{5{{z}_{2}}}{11{{z}_{1}}}=iy $ (say) or $ \frac{{{z}_{2}}}{{{z}_{1}}}=\frac{11}{5}iy $
Now, $ \left| \frac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=\frac{\left| 2+3\frac{{{z}_{2}}}{{{z}_{1}}} \right|}{\left| 2-3\frac{{{z}_{2}}}{{{z}_{1}}} \right|} $
$=\frac{\left| 2+\frac{33}{5}iy \right|}{\left| 2-\frac{33}{5}iy \right|}=1 $
Now, $ \left| \frac{2{{z}_{1}}+3{{z}_{2}}}{2{{z}_{1}}-3{{z}_{2}}} \right|=\frac{\left| 2+3\frac{{{z}_{2}}}{{{z}_{1}}} \right|}{\left| 2-3\frac{{{z}_{2}}}{{{z}_{1}}} \right|} $
$=\frac{\left| 2+\frac{33}{5}iy \right|}{\left| 2-\frac{33}{5}iy \right|}=1 $
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
