Let O be the origin and A be the point z1 = 1 + 2i. If B is the point z2, Re(z2) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

z2−0 (1+2i)−0=|OBOA|eiπ4⇒z2= (1+2i) (1+i)=−1+3i∴arg z2=π−tan−13 and |z2|=10z1−2z2=3−4i∴arg (z1−2z2)=−tan−143⇒|z1−2z2|=|2+4i+1−3i|=10

Let O be the origin and A be the point z1 = 1 + 2i. If B is the point z2, Re(z2) &lt; 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

arg z2 π - tan-3 3

arg(z1 – 2z2) = tan-1 <math><mrow><mfrac><mrow><mn>4</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mrow></math>

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Maths Class 12th Complex Numbers and Quadratic Equations Maths NCERT Exemplar Solutions Class 12th Chapter One

Let O be the origin and A be the point z₁ = 1 + 2i. If B is the point z₂, Re(z₂) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

Option 1 -

arg z₂ π - tan^-3 3

Option 2 -

arg(z₁ – 2z₂) = tan-1 $\frac{4}{3}$

Option 3 -

$| z_{2} | = \sqrt[]{10}$

Option 4 -

$| 2 z_{1} - z_{2} | = 5$

2 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$\frac{z_{2} - 0}{(1 + 2 i) - 0} = | \frac{O B}{O A} | e^{\frac{i π}{4}} \Rightarrow z_{2} = (1 + 2 i) (1 + i) = - 1 + 3 i ∴ a r g z_{2} = π - t a n^{- 1} 3 a n d | z_{2} | = \sqrt[]{10}$

$z_{1} - 2 z_{2} = 3 - 4 i ∴ a r g (z_{1} - 2 z_{2}) = - t a n^{- 1} \frac{4}{3} \Rightarrow | z_{1} - 2 z_{2} | = | 2 + 4 i + 1 - 3 i | = \sqrt[]{10}$

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Vishal Baghel

(2 – i) z = (2 + i) $\bar{z}$ , put z = x + iy

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A quadratic function f(x) satisfies f(x) ≥ 0 for all real x. If f(2) = 0 and f(4) = 12, then the value of f(6) is :

Vishal Baghel

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 × 4² = 48

The area of the triangle vertices A(z), B(iz) and C(z + iz) is:

Vishal Baghel

Kindly consider the following figure

Let z be those complex numbers which satisfy

$| z + 5 | \leq 4 a n d z (1 + i) + \bar{z} (1 - i) \geq - 10, i = \sqrt[]{- 1} .$

If the maximum value of ${| z + 1 |}^{2} i s α + β \sqrt[]{2},$ then the value of (α + β) is…….

Vishal Baghel

$G i v e n | z + 5 | \leq 4 . . . . . . . . . . (i)$

->Represent a circle ${(x + 5)}^{2} + y^{2} \leq 16$

$= z (1 + i) + \bar{z} (1 - i) \geq - 10 . . . . . . . . . . (i i)$

->Represent a line X – y $\geq - 5$

So max |z + 1|² = AQ²

$= {(- 4 - 2 \sqrt[]{2})}^{2} + 8$

$\begin{array}{l} = 32 + 16 \sqrt[]{2} \\ = α + β \sqrt[]{2} \end{array}$

Hence α + β) = 48

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Vishal Baghel

$S = 21 + \sum_{k = 1}^{21} {(z^{k} + \frac{1}{z^{k}})}^{3} = \sum {(z^{k} + {\bar{z}}^{k})}^{3} = 8 \sum {(R e (z^{k}))}^{3}$

$(A s | z | = 1 \Rightarrow \frac{1}{z} = \bar{z})$

$S = 2 \sum_{k = 1}^{21} c o s k π + 6 \sum_{k = 1}^{21} c o s \frac{k π}{3} + 21$

= $2 (- 1) + 6 R e (α + α^{2} + α^{3}) + 21$

Where $α = e^{i (2 π / 6)}$

$= 2 (- 1) + 6 (\frac{1}{2} - \frac{1}{2} - 1) + 21$

$= - 8 + 21 = 13$

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Let O be the origin and A be the point z₁ = 1 + 2i. If B is the point z₂, Re(z₂) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

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