If arg z1-z-z-z-z-2 and -z-z-z1-3 then -z1-2 equals to

( z_1 z|z|z|z| 0)=π2 ⇒( 1)+( z|z| z_1z|z| 0)=π2 ⇒( z|z| z_1z|z| 0)= π2 So, locus of z|z| will be as shown below Given nbsp; nbsp; nbsp;| z|z| z_1|=3 ⇒ PA ...

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Standard XII

Mathematics

Rotation Concept

If arg z1-z/|...

Question

If $arg ⎛ ⎜ ⎜ ⎜ ⎝ \frac{z_{1} - \frac{z}{| z |}}{\frac{z}{| z |}} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{π}{2}$ and $∣ ∣ ∣ \frac{z}{| z |} - z_{1} ∣ ∣ ∣ = 3$ then $| z_{1} |^{2}$ equals to

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Solution

$arg ⎛ ⎜ ⎜ ⎜ ⎝ \frac{z_{1} - \frac{z}{| z |}}{\frac{z}{| z |} - 0} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{π}{2}$
$\Rightarrow arg (- 1) + arg ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{z}{| z |} - z_{1}}{\frac{z}{| z |} - 0} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{π}{2}$
$\Rightarrow arg ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{z}{| z |} - z_{1}}{\frac{z}{| z |} - 0} ⎞ ⎟ ⎟ ⎟ ⎠ = - \frac{π}{2}$
So, locus of $\frac{z}{| z |}$ will be as shown below

Given $∣ ∣ ∣ \frac{z}{| z |} - z_{1} ∣ ∣ ∣ = 3$
$\Rightarrow P A = 3$
and $P B = ∣ ∣ ∣ \frac{z}{| z |} ∣ ∣ ∣ = 1$
$\Rightarrow A B = \sqrt{(P A)^{2} + (P B)^{2}} = \sqrt{10}$
$\Rightarrow | z_{1} |^{2} = 10$

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Rotation Concept

Standard XII Mathematics

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If arg⎛⎜ ⎜ ⎜⎝z1−z|z|z|z|⎞⎟ ⎟ ⎟⎠=π2 and ∣∣∣z|z|−z1∣∣∣=3 then |z1|2 equals to

If $arg ⎛ ⎜ ⎜ ⎜ ⎝ \frac{z_{1} - \frac{z}{| z |}}{\frac{z}{| z |}} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{π}{2}$ and $∣ ∣ ∣ \frac{z}{| z |} - z_{1} ∣ ∣ ∣ = 3$ then $| z_{1} |^{2}$ equals to