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Question: Let ABC be an isosceles triangle inscribed in the circle |z| = r with AB = AC . if z1 , z2 , z3 repr...

Let ABC be an isosceles triangle inscribed in the circle |z| = r with AB = AC . if z1 , z2 , z3 represent A , B , C then find relation b/w them

Answer

z_1^2 = z_2 z_3

Explanation

Solution

Let the circle be centered at the origin O in the complex plane with radius r. The equation of the circle is ∣z∣=r|z|=r. The vertices of the triangle A, B, C are represented by the complex numbers z1,z2,z3z_1, z_2, z_3 respectively, and they lie on the circle. Thus, ∣z1∣=∣z2∣=∣z3∣=r|z_1|=|z_2|=|z_3|=r. This implies zkzkˉ=r2z_k \bar{z_k} = r^2, so zkˉ=r2/zk\bar{z_k} = r^2/z_k for k=1,2,3k=1, 2, 3.

The triangle ABC is isosceles with AB = AC. The distance between two points represented by complex numbers zaz_a and zbz_b is ∣za−zb∣|z_a - z_b|. Given AB = AC, we have ∣z1−z2∣=∣z1−z3∣|z_1 - z_2| = |z_1 - z_3|. Squaring both sides, we get ∣z1−z2∣2=∣z1−z3∣2|z_1 - z_2|^2 = |z_1 - z_3|^2. Using the property ∣w∣2=wwˉ|w|^2 = w \bar{w}, we have: (z1−z2)(z1−z2‾)=(z1−z3)(z1−z3‾)(z_1 - z_2)(\overline{z_1 - z_2}) = (z_1 - z_3)(\overline{z_1 - z_3}) (z1−z2)(z1ˉ−z2ˉ)=(z1−z3)(z1ˉ−z3ˉ)(z_1 - z_2)(\bar{z_1} - \bar{z_2}) = (z_1 - z_3)(\bar{z_1} - \bar{z_3}) z1z1ˉ−z1z2ˉ−z2z1ˉ+z2z2ˉ=z1z1ˉ−z1z3ˉ−z3z1ˉ+z3z3ˉz_1 \bar{z_1} - z_1 \bar{z_2} - z_2 \bar{z_1} + z_2 \bar{z_2} = z_1 \bar{z_1} - z_1 \bar{z_3} - z_3 \bar{z_1} + z_3 \bar{z_3}

Since ∣z1∣=r|z_1|=r, ∣z2∣=r|z_2|=r, ∣z3∣=r|z_3|=r, we have z1z1ˉ=r2z_1 \bar{z_1} = r^2, z2z2ˉ=r2z_2 \bar{z_2} = r^2, z3z3ˉ=r2z_3 \bar{z_3} = r^2. Substituting these into the equation: r2−z1z2ˉ−z2z1ˉ+r2=r2−z1z3ˉ−z3z1ˉ+r2r^2 - z_1 \bar{z_2} - z_2 \bar{z_1} + r^2 = r^2 - z_1 \bar{z_3} - z_3 \bar{z_1} + r^2 −z1z2ˉ−z2z1ˉ=−z1z3ˉ−z3z1ˉ- z_1 \bar{z_2} - z_2 \bar{z_1} = - z_1 \bar{z_3} - z_3 \bar{z_1} z1z2ˉ+z2z1ˉ=z1z3ˉ+z3z1ˉz_1 \bar{z_2} + z_2 \bar{z_1} = z_1 \bar{z_3} + z_3 \bar{z_1}

Now, substitute zkˉ=r2/zk\bar{z_k} = r^2/z_k for k=1,2,3k=1, 2, 3: z1(r2/z2)+z2(r2/z1)=z1(r2/z3)+z3(r2/z1)z_1 (r^2/z_2) + z_2 (r^2/z_1) = z_1 (r^2/z_3) + z_3 (r^2/z_1) Divide by r2r^2 (since r≠0r \neq 0): z1/z2+z2/z1=z1/z3+z3/z1z_1/z_2 + z_2/z_1 = z_1/z_3 + z_3/z_1 Multiply by z1z2z3z_1 z_2 z_3 to clear the denominators (since z1,z2,z3≠0z_1, z_2, z_3 \neq 0 for vertices of a triangle inscribed in a circle centered at the origin): z12z3+z22z3=z12z2+z32z2z_1^2 z_3 + z_2^2 z_3 = z_1^2 z_2 + z_3^2 z_2 Rearrange the terms: z12z3−z12z2+z22z3−z32z2=0z_1^2 z_3 - z_1^2 z_2 + z_2^2 z_3 - z_3^2 z_2 = 0 Factor by grouping: z12(z3−z2)+z2z3(z2−z3)=0z_1^2 (z_3 - z_2) + z_2 z_3 (z_2 - z_3) = 0 z12(z3−z2)−z2z3(z3−z2)=0z_1^2 (z_3 - z_2) - z_2 z_3 (z_3 - z_2) = 0 (z12−z2z3)(z3−z2)=0(z_1^2 - z_2 z_3)(z_3 - z_2) = 0

Since B and C are distinct vertices of a triangle, z2≠z3z_2 \neq z_3, so z3−z2≠0z_3 - z_2 \neq 0. Therefore, we must have z12−z2z3=0z_1^2 - z_2 z_3 = 0. z12=z2z3z_1^2 = z_2 z_3.

This is the relation between z1,z2,z3z_1, z_2, z_3 when A is the apex and AB = AC.