Let S 1 = { z ? C : | z | ? 5 } , S 2 = { z ? C : Im ( z + 1 - 3 i 1 - 3 i ) ? 0 } and S 3 = { z ? C : Re ( z ) ? 0 } . Then the area of the region S 1 ? S 2 ? S 3 is : [2024]

Question

Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

Smart Achievers · Accepted Answer

(1)

$Let z = x + i y be any complex number$

$S_{1} = {z \in C : | z | \leq 5}$

$S_{1} : x^{2} + y^{2} \leq 25$ ...(i)

$S_{2} : Im [\frac{x + iy}{1 - \sqrt{3} i} + 1] \geq 0$

$i.e.,$ $S_{2} : Im [\frac{(x + iy) (1 + \sqrt{3} i)}{4} + 1] \geq 0$

$\Rightarrow \sqrt{3} x + y \geq 0$ ...(ii)

$S_{3} : {z \in C : Re (z) \geq 0}$

$\Rightarrow x \geq 0$ ...(iii)

$Now,$ $\sqrt{3} x + y = 0$

$\Rightarrow y = - \sqrt{3} x$

$\Rightarrow \tan θ = 120 °$

$∴$ $Required area = \frac{25 π}{2} - \frac{1}{12} {(5)}^{2} π$

$= \frac{25 π}{2} - \frac{25 π}{12}$

$= \frac{125 π}{12}$

Solution

Q.
Let $S_{1} = {z \in C : | z | \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - \sqrt{3} i}{1 - \sqrt{3} i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0} .$ Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is : [2024]

1 $\frac{125 π}{12}$

2 $\frac{125 π}{24}$

3 $\frac{125 π}{6}$

4 $\frac{125 π}{4}$

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Solution

Q. Let S1={z∈C:|z|≤5}, S2={z∈C:Im(z+1-3i1-3i)≥0} and S3={z∈C:Re(z)≥0}. Then the area of the region S1∩S2∩S3 is : [2024]

1 125π12

2 125π24

3 125π6

4 125π4