Let the curve , divide the region into two parts of area and . Then equals : [2025]

Question

Let the curve $z (1 + i) + \bar{z} (1 - i) = 4, z \in C$ , divide the region $| z - 3 | \leq 1$ into two parts of area $α$ and $β$ . Then $| α - β |$ equals : [2025]

Smart Achievers · Accepted Answer

(4)

Let z = x + iy, then from given equation, we have

(x + iy)(1 + i) + (x – iy)(1 – i) = 4

$\Rightarrow$ x + ix + iy – y + x – ix – iy – y = 4

$\Rightarrow$ 2x – 2y = 4 $\Rightarrow$ x – y = 2

Now, $| z - 3 | \leq 1 \Rightarrow {(x - 3)}^{2} + y^{2} \leq 1$

$β$ = Area of shaded region = $\frac{π {(1)}^{2}}{4} - \frac{1}{2} \times 1 \times 1$

= $(\frac{π}{4} - \frac{1}{2})$ sq. units

$α$ = Area of unshaded region inside the circle

$= \frac{3}{4} π {(1)}^{2} + \frac{1}{2} \times 1 \times 1 = (\frac{3 π}{4} + \frac{1}{2})$ sq. units

Now, $| α - β |$ = difference of area

= $(\frac{3 π}{4} + \frac{1}{2}) - (\frac{π}{4} - \frac{1}{2}) = \frac{π}{2} + 1$ .

Solution

Q.
Let the curve $z (1 + i) + \bar{z} (1 - i) = 4, z \in C$ , divide the region $| z - 3 | \leq 1$ into two parts of area $α$ and $β$ . Then $| α - β |$ equals : [2025]

1 $1 + \frac{π}{3}$

2 $1 + \frac{π}{6}$

3 $1 + \frac{π}{4}$

4 $1 + \frac{π}{2}$

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Solution

Q. Let the curve z(1+i)+z¯(1–i)=4, z∈C, divide the region |z–3|≤1 into two parts of area α and β. Then |α–β| equals : [2025]

1 1+π3

2 1+π6

3 1+π4