True False[1 Marks ]

Question 11 Mark

The locus represented by| z - 1| = |z - i| is a line perpendicular to the join of (1, 0) and (0, 1).

Answer

True.
Solution:
We have, |z - 1| = |z - i|
Putting z = x + iy, we get
⇒ |x - 1 + iy| = |x - i(1 - y)|
⇒ (x - 1)² + y² = x² + (1 - y)²
⇒ x² - 2x + 1 + y² = x² + 1 + y² - 2y
⇒ -2x + 1 = 1 - 2y
⇒ -2x + 2y = 0
⇒ x - y = 0
Now, equation of a line through the points (1, 0) and (0, 1) is,
$\text{y}-0=\frac{1-0}{0-1}(\text{x}-1)$
Or x + y = 1
This line is perpendicular to the line x - y = 0

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Question 21 Mark

Multiplication of a non zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.

Answer

False.
Solution:
Let z = x + iy, where x, y > 0
i.e., z or point A(x, y) lies in first quadrant. Now, -iz = -i(x + iy)
= -ix - i²y = y - ix
Now, point B(y, -x) lies in fourth quadrant. Also, ∠AOB = 90°
Thus, B is obtained by rotating A in clockwise direction about origin.

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Question 31 Mark

2 is not a complex number.

Answer

False.
Solution:
We know that, any real number is also a complex number.

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Question 41 Mark

The order relation is defined on the set of complex numbers.

Answer

False.
Solution:
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

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Question 51 Mark

If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z²) = 0.

Answer

False.
Solution:
Let z = x + iy, z ≠ 0 and Re(z) = 0
i.e., x = 0
$\therefore$ z = iy
Im(z²) = i²y² = -y² ≠ 0

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Question 61 Mark

The inequality |z - 4| < |z - 2| represents the region given by x > 3.

Answer

True.
Solution:
We have, |z - 4| < |z - 2|
Putting z = x + iy, we get
$|\text{x}-4+\text{iy}|<|\text{x}-2+\text{iy}|$
$\Rightarrow\sqrt{(\text{x}-4)^2+\text{y}^2}<\sqrt{(\text{x}-2)^2+\text{y}^2}$
$\Rightarrow(\text{x}-4)^2+\text{y}^2<(\text{x}-2)^2+\text{y}^2$
$\Rightarrow\text{x}^2-8\text{x}+16+\text{y}^2<\text{x}^2-4\text{x}+4+\text{y}^2$
$\Rightarrow-8\text{x}+16<-4\text{x}+4$
$\Rightarrow4\text{x}>12$
$\Rightarrow\text{x}>3$

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Question 71 Mark

Let z₁ and z₂ be two complex numbers such that |z₁ + z₂| = |z₁| + |z₂|, then arg(z₁ - z₂) = 0.

Answer

False.
Solution:
$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
$\Rightarrow|\text{z}_1+\text{z}_2|^2=|\text{z}_1|^2+|\text{z}_2|^2+2|\text{z}_1||\text{z}_2|$
$\Rightarrow|\text{z}_1|^2+|\text{z}_2|^2+2\text{Re}(\text{z}_1\bar{\text{z}}_2)=|\text{z}_1|^2+|\text{z}_2|^2+2|\text{z}_1||\text{z}_2|$
$\Rightarrow2\text{Re}(\text{z}_1\bar{\text{z}}_2)=2|\text{z}_1||\text{z}_2|$
$\Rightarrow\cos(\theta_1-\theta_2)=1$
$\Rightarrow\theta_1-\theta_2=0$
$\Rightarrow\arg(\text{z}_1)-\arg(\text{z}_2)=0$

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Question 81 Mark

For any complex number z the minimum value of |z| + | z - 1| is 1.

Answer

True.
Solution:
We know that $|\text{z}_1|+|\text{z}_2|\geq|\text{z}_1-\text{z}_1|$
$\Rightarrow|\text{z}|+|\text{z}-1|\geq|\text{z}-(\text{z}-1)|$
$\Rightarrow|\text{z}|+|\text{z}-1|\geq1$
So, minimum value of |z| + |z - 1| is 1.
Alternative method
Let A(z) and B(1)
$\Rightarrow|\text{z}|+\text{z}-1|=\text{OA}+\text{AB},$ where O is origin
From triangular inequality, we get
$\text{OA}+\text{AB}\geq\text{OB}$
$\Rightarrow(\text{OA}+\text{AB})_{\text{min}}=\text{OB}=1$

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