If $a$ and $b$ are real numbers between $0$ and $1$ such that points $z_{1} = 1 + i b$ and $z_{2} = a + i$ and $z_{3} = 0$ form an ,equilateral triangle then ordered pair (a, b) is given by

(a, a)

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(b, a)

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(b, b)

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All of these

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Solution

The correct option is D All of these
Since the following vertices, form an equilateral triangle.
$z_{1}^{2} + z_{2}^{2} + 0^{2} = z_{1} (0) + z_{2} (0) + z_{1} . z_{2}$
$(1 + i b)^{2} + (a + i)^{2} = (1 + i b) (a + i)$
$1 - b^{2} + i 2 b + a^{2} - 1 + i 2 a = 1 + i + i (a b) - 1$
$a^{2} - b^{2} + i 2 (a + b) = i (1 + a b)$
Therefore
$a^{2} - b^{2} = 0$
$a^{2} = b^{2}$
$a = \pm b$

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Similar questions

z_{1} = a - i, z_{2} = - 1 + i b

z_{3} = 0

z_{1} = a + i, z_{2} = b + i a n d z_{3} = 0

z_{1} = a + i, z_{2} = 1 + b i

z_{3} = 0

a

b

z_{1}, z_{2}, z_{3}

z_{2} \neq 1, a = | z_{1} |, b = | z_{2} |

c = | z_{3} |

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

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