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Maths Fill In The Blanks[1 Marks ]

Complex Numbers and Quadratic Equations · CBSE STD 11 Science (CBSE - English Medium). 10 questions.

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Fill In The Blanks[1 Marks ]

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Question 11 Mark
Fill in the blanks.
The value of $\sqrt{-25}\times\sqrt{-9}$ is ___________.
Answer
The value of $\sqrt{-25}\times\sqrt{-9}$ is -15.
Solution:
$\sqrt{-25}\times\sqrt{-9}=\text{i}\sqrt{25}\times\text{i}\sqrt{9}$
$=\text{i}^2(5\times3)=-15$
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Question 21 Mark
Fill in the blanks.
For any two complex numbers $z_1 , z_2 $ and any real numbers $a, b, |az_1 - bz_2|^2 + |bz_1 + az_2|^2= \_\_\_\_\_\_\_\_\_\_\_.$
Answer
For any two complex numbers $z_1 , z_2 $ and any real numbers $a, b, |az_1 - bz_2|^2 + |bz_1 + az_2|^2 =(a^2 + b^2)(|z_1|^2 + |z_2|^2).$​​​​​​​
Solution:
$|\text{az}_1-\text{bz}_2|^2+|\text{bz}_1+\text{az}_2|^2$
$=|\text{az}_1|^2+|\text{bz}_2|^2-2\text{Re}(\text{az}_1\times\text{b}\bar{\text{z}}_2)+|\text{bz}_1|^2+|\text{az}_2|^2+2\text{Re}(\text{bz}_1\times\text{a}\bar{\text{z}}_2)$
$=(\text{a}^2+\text{b}^2)|\text{z}_1|^2+(\text{a}^2+\text{b}^2)|\text{z}_2|^2=(\text{a}^2+\text{b}^2)(|\text{z}_1|^2+|\text{z}_2|^2)$
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Question 31 Mark
Fill in the blanks.
The sum of the series $i + i^2 + i^3 + ...$ upto $1000$ terms is __________.
Answer
The sum of the series $i + i^2 + i^3 + ...$ upto $1000$ terms is $0$.
Solution:
$i + i^2 + i^3 + ...$ upto $1000$ terms.
$= (i + i^2 + i^3 + i^4) + (i5 + i6 + i7 + i8) + ....... 250$ brackets.
$= 0 + 0 + 0 + ..... + 0 \big[\because i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0,$
where $\text{n}\in\text{N}\big]$
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Question 41 Mark
Fill in the blanks.
$\arg(\text{z})+\arg\bar{\text{z}}(\bar{\text{z}}\neq0)\text{ is}$ _________.
Answer
$\arg(\text{z})+\arg\bar{\text{z}}(\bar{\text{z}}\neq0)\text{ is}$ 0.Solution:
$\arg(\text{z})+\arg\bar{\text{z}}=\theta+(-\theta)=0$
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Question 51 Mark
Fill in the blanks.
If $\Big|\frac{\text{z}-2}{\text{z}+2}\Big|=\frac{\pi}{6},$ then the locus of z is __________.
Answer
If $\Big|\frac{\text{z}-2}{\text{z}+2}\Big|=\frac{\pi}{6},$ then the locus of z is 0.Solution:
$\Rightarrow\frac{|\text{x}+\text{iy}-2|}{|\text{x}+\text{iy}+2|}=\frac{\pi}{6}$ $\Rightarrow\frac{|\text{x}-2+\text{iy}|}{|\text{x}+2+\text{iy}|}=\frac{\pi}{6}$ $\Rightarrow6|\text{x}-2+\text{iy}|=\pi|\text{x}+2+\text{iy}|$ $\Rightarrow36|\text{x}-2+\text{iy}|^2=\pi^2|\text{x}+2+\text{iy}|^2$ $\Rightarrow36[\text{x}^2-4\text{x}+4+\text{y}^2\big]=\pi^2[\text{x}^2+4\text{x}+4+\text{y}^2]$ $\Rightarrow(36-\pi^2)\text{x}^2+(36-\pi^2)\text{y}^2-(144+4\pi^2)\text{x}+144-4\pi^2=0,$ which is a circle.
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Question 61 Mark
Fill in the blanks.
The number $\frac{(1-\text{i})^3}{1-\text{i}^3}$ is equal to __________.
Answer
The number $\frac{(1-\text{i})^3}{1-\text{i}^3}$ is equal to -2.Solution:
$\frac{(1-\text{i})^3}{1-\text{i}^3}=\frac{(1-\text{i})^3}{(1-\text{i})(1+\text{i}+\text{i}^2)}=\frac{(1-\text{i})^2}{\text{i}}$ $=\frac{1+\text{i}^2-2\text{i}}{\text{i}}=\frac{-2\text{i}}{\text{i}}=-2$
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Question 71 Mark
Fill in the blanks.
If $z_1 $ and $z_2 $ are complex numbers such that $z_1 + z_2 $ is a real number, then $z_2 = \_\_\_\_\_\_\_\_\_\_\_\_\_\_.$
Answer
If $z_1 $ and $z_2 $ are complex numbers such that $z_1 + z_2 $ is a real number, then $z_2 = \bar{\text{z}}_1$
Solution:
Let $z_1 = x_1 + iy_1 $ and $z_2 = x_2 + iy_2 $
$z_1+ z_2= (x_1 + x_2) + i(y_1 + y_2),$
which is real $\Rightarrow y_1 + y_2 = 0 $
$\Rightarrow y_1 = -y_2$_
$\because  z_2 = x_1 - iy_1 [$Assuming $x_1 = x_2]$
$\Rightarrow\text{z}_2=\bar{\text{z}}_1$
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Question 81 Mark
Fill in the blanks.
If z = 4 and $\arg(\text{z})=\frac{5\pi}{6},$ then z = __________.
Answer
If z = 4 and $\arg(\text{z})=\frac{5\pi}{6},$ then z $=-2\sqrt{3}+2\text{i}$Solution:
Let $\text{z}=|\text{z}|(\cos\theta+\text{i}\sin\theta)$
Where $\theta=\text{arg(z)}$
Given that |z| = 4 and $\text{arg(z)}=\frac{5\pi}{6}$
$\text{z}=4\Big[\cos\frac{5\pi}{6}+\text{i}\sin\frac{5\pi}{6}\Big]$ (z lies in II quadrant)
$=4\Big[-\frac{\sqrt{3}}{2}+\text{i}\frac{1}{2}\Big]$
$=-2\sqrt{3}+2\text{i}$
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Question 91 Mark
Fill in the blanks.
Multiplicative inverse of 1 + i is _________.
Answer
Multiplicative inverse of 1 + i is $\frac{1}{2}(1-\text{i})$Solution:
Multiplicative inverse of $1+\text{i}$
$\frac{1}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}=\frac{1-\text{i}}{1+1}=\frac{1-\text{i}}{2}=\frac{1}{2}(1-\text{i})$
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Question 101 Mark
Fill in the blanks.
If $|\text{z}+4|\leq3,$ then the greatest and least values of z + 1 are ..... and ____________.
Answer
If $|\text{z}+4|\leq3,$ then the greatest and least values of z + 1 are ..... and 0.
Solution:
Given that, $|\text{z}+4|\leq3,$
For the greatest value of |z + 1|,
$|\text{z}+1|=|\text{z}+4-3|$
Or $|\text{z}+1|\leq|\text{z}+4|+|-3|$
Or $|\text{z}+1|\leq3+3$
Or $|\text{z}+1|\leq6$
So, greatest value of |z + 1| is 6.
We know that the least value of the modulus of a complex number is zero.
So, the least value of |z + 1| is zero.
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