True False[1 Marks ]

Question 11 Mark

Let $z_1$ and $z_2$ be two complex numbers such that $|z_1 + z_2| = |z_1| + |z_2|,$ then arg $(z_1 - z_2) = 0.$

Answer

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

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Question 21 Mark

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 31 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 41 Mark

If $z$ is a complex number such that $z \neq 0$ and Re $(z) = 0,$ then $Im (z^2) = 0.$

Answer

False.
Let $z = x + iy, z \neq 0$ and $Re(z) = 0$
$i.e., x = 0$
$\therefore z = iy$
$Im(z^2) = i^2y^2 = -y^2 \neq 0$

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Question 51 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 61 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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Question 71 Mark

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

Answer

True.
We have$, |z - 1| = |z - i|$
Putting$ z = x + iy,$ we get
$\Rightarrow |x - 1 + iy| = |x - i(1 - y)|$
$\Rightarrow (x - 1)^2 + y^2 = x^2 + (1 - y)^2$
$\Rightarrow x^2 - 2x + 1 + y^2 = x^2 + 1 + y^2 - 2y$
$\Rightarrow -2x + 1 = 1 - 2y$
$\Rightarrow -2x + 2y = 0$
$\Rightarrow x - y = 0$
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 81 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.
Let $ z = x + iy,$ where $x, y > 0$
$i.e., z$ or point $A(x, y)$ lies in first quadrant. Now$, -iz = -i(x + iy)$
$= -ix - i^2y = y - ix$
Now$,$ point $B(y, -x)$ lies in fourth quadrant. Also$, \angle AOB = 90^\circ$
Thus$, B$ is obtained by rotating $A$ in clockwise direction about origin.

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