JEE Advanced · Mathematics · 3. Complex Numbers
If \(z\) is any complex number satisfying \(|z-3-2 i| \leq 2\), then the maximum value of \(|2 z-6+5 i|\) is
- A 5
- B 10
- C 15
- D 20
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
Given, \(|z-3-2 i| \leq 2\)
To find minimum of \(|2 z-6+5 i|\) or \(\quad 2\left|z-3+\frac{5}{2} i\right|\)
Using triangle inequality, i.e. ||\(z_1|-| z_2|| \leq\left|z_1+z_2\right|\) \(\therefore\left|z-3+\frac{5}{2} i\right|\) \(=\left|z-3-2 i+2 i+\frac{5}{2} i\right|\) \(=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|\) \(\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}\) or \(\quad|2 z-6+5 i| \geq 5\)
To find minimum of \(|2 z-6+5 i|\) or \(\quad 2\left|z-3+\frac{5}{2} i\right|\)
Using triangle inequality, i.e. ||\(z_1|-| z_2|| \leq\left|z_1+z_2\right|\) \(\therefore\left|z-3+\frac{5}{2} i\right|\) \(=\left|z-3-2 i+2 i+\frac{5}{2} i\right|\) \(=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|\) \(\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}\) or \(\quad|2 z-6+5 i| \geq 5\)
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