SATHEE: Complex Numbers 3 Question 11

Complex Numbers 3 Question 11

11. $| z | \leq 1, | w | \leq 1$ , then show that

$| z - w |^{2} \leq (| z | - | w |)^{2} + (\arg z - \arg w)^{2}$

(1995, 5M)

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Solution:

Let $z = r_{1} (\cos θ_{1} + i \sin θ_{1})$ and $w = r_{2} (\cos θ_{2} + i \sin θ_{2})$ We have, $| z | = r_{1}, | w | = r_{2}, \arg (z) = θ_{1}$ and $\arg (w) = θ_{2}$ Given, $| z | \leq 1, | w | < 1$

$\Rightarrow r_{1} \leq 1$ and $r_{2} \leq 1$

Now,

$\begin{matrix} z - w = (r_{1} \cos θ_{1} - r_{2} \cos θ_{2}) + i (r_{1} \sin θ_{1} - r_{2} \sin θ_{2}) \\ \Rightarrow | z - w |^{2} = {(r_{1} \cos θ_{1} - r_{2} \cos θ_{2})}^{2} + {(r_{1} \sin θ_{1} - r_{2} \sin θ_{2})}^{2} \\ = r_{1}^{2} \cos^{2} θ_{1} + r_{2}^{2} \cos^{2} θ_{2} - 2 r_{1} r_{2} \cos θ_{1} \cos θ_{2} \\ + r_{1}^{2} \sin^{2} θ_{1} + r_{2}^{2} \sin^{2} θ_{2} - 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2} \\ = r_{1}^{2} (\cos^{2} θ_{1} + \sin^{2} θ_{1}) + r_{2}^{2} (\cos^{2} θ_{2} + \sin^{2} θ_{2}) \\ - 2 r_{1} r_{2} (\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2}) \\ = r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) \\ = {(r_{1} - r_{2})}^{2} + 2 r_{1} r_{2} [1 - \cos (θ_{1} - θ_{2})] \\ = {(r_{1} - r_{2})}^{2} + 4 r_{1} r_{2} \sin^{2} \frac{θ_{1} - θ_{2}}{2} \\ \leq {| r_{1} - r_{2} |}^{2} + 4 \sin \frac{θ_{1} - θ_{2}}{2}, r_{2} \leq 1] \end{matrix}$

and $| \sin θ | \leq | θ |, \forall θ \in R$

Therefore, $| z - w |^{2} \leq {| r_{1} - r_{2} |}^{2} + 4 \frac{θ_{1} - θ_{2}}{2}$

$\leq {| r_{1} - r_{2} |}^{2} + {| θ_{1} - θ_{2} |}^{2}$

$\Rightarrow | z - w |^{2} \leq (| z | - | w |)^{2} + (\arg z - \arg w)^{2}$

Alternate Solution

$\begin{matrix} | z - w |^{2} = | z |^{2} + | w |^{2} - 2 | z | | w | \cos (\arg z - \arg w) \\ = | z |^{2} + | w |^{2} - 2 | z | | w | + 2 | z | | w | \\ - 2 | z | | w | \cos (\arg z - \arg w) \\ = (| z | - | w |)^{2} + 2 | z | | w | \cdot 2 \sin^{2} \frac{\arg z - \arg w}{2} \dots (i) \\ ∴ | z - w |^{2} \leq (| z | - | w |)^{2} + 4 \cdot 1 \cdot 1 \frac{\arg z - \arg w^{2}}{2} \\ \Rightarrow | z - w |^{2} \leq (| z | - | w |)^{2} + (\arg z - \arg w)^{2} \end{matrix}$

Complex Numbers 3 Question 11

11. $| z | \leq 1, | w | \leq 1$ , then show that

Solution:

Alternate Solution

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Complex Numbers 3 Question 11

11. |z|≤1,|w|≤1, then show that

Solution:

Alternate Solution

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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11. $| z | \leq 1, | w | \leq 1$ , then show that