True False[1 Marks ]

Question 11 Mark

State True or False for the following:
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 21 Mark

State True or False for the following:
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 31 Mark

State True or False for the following:
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 41 Mark

State True or False for the following:
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 51 Mark

State True or False for the following:
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 61 Mark

State True or False for the following:
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 71 Mark

State True or False for the following:
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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Question 81 Mark

State True or False for the following:
2 is not a complex number.

Answer

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

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MATHS True False[1 Marks ]

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