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Locus of z when (z - a)/(z - b) is purely real or purely imaginary

This type of question, asked 5 times: 2021 Paper 20, 2021 Paper 23, 10 Apr 2023, 9 Apr 2024, 7 Apr 2025. All 6 questions below, oldest first.

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  1. Question 1

    2021 Paper 20

    67
    If S={z∈C:z−iz+2i∈R}S=\left\{ z\in\mathbb{C}:\frac{z -i}{z + 2i}\in\mathbb{R} \right\}, then:

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  2. Question 2

    2021 Paper 23

    74
    If zz is a complex number such that z−iz−1\frac{z - i}{z - 1} is purely imaginary, then the minimum value of ∣z−(3+3i)∣|z- (3 + 3i)| is :

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  3. Question 3

    10 Apr 2023

    67
    Let the complex number z=x+z = x + iy be such that 2z−3i2z+i\frac{2z - 3i}{2z + i} is purely imaginary. If x+y2=0x +y^{2}= 0, then y4+y2−yy^{4}+y^{2}- y is equal to:

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  4. Question 4

    10 Apr 2023

    168
    Let S={z=x+iy:2z−3i4z+2iS =\left\{ z = x + iy:\frac{2z - 3i}{4z + 2i} \right. is a real number }\}. Then which of the following is NOT correct?

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  5. Question 5

    9 Apr 2024

    154
    Let zz be a complex number such that the real part of z−2iz+2i\frac{z- 2i}{z+ 2i} is zero. Then, the maximum value of ∣z−(6+8i)∣|z - (6 + 8i)| is equal to:

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  6. Question 6

    7 Apr 2025

    61
    Among the statements (S1): The set {z∈C−{−i}:∣z∣=1 and z−iz+i is purely real}\{z\in\mathbb{C}-\{-i\}:|z|=1\text{ and }\frac{z-i}{z+i}\text{ is purely real}\} contains exactly two elements, and (S2): The set {z∈C−{−1}:∣z∣=1 and z−1z+1 is purely imaginary}\{z\in\mathbb{C}-\{-1\}:|z|=1\text{ and }\frac{z-1}{z+1}\text{ is purely imaginary}\} contains infinitely many elements.

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