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CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Number Theory
Quiz 12
The modulus of the complex number $$z$$ such that $$\left| z + 3 - i\right | = 1$$ and $$\arg{z} = \pi$$ is equal to
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$$1$$
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$$2$$
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$$4$$
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$$3$$
Explanation
$$arg(z)=\pi$$
Let $$z=x+iy$$
$$arg(x+iy)=\pi$$
$$\Rightarrow x < 0; y=0$$
$$|z+3-i|=1$$
$$|x+3-i|=1$$
$$(x+3)^2+1^2=1^2$$
$$\Rightarrow (x+3)^2=0$$
$$x=-3$$
$$|z|=|-3+0i|=3$$.
The roots of the equation $$(3b+c-4a)x^2+(3c+a-4b)x+(3a+b-4c)= 0$$ are
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Irrational
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Rational
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Non-real
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Imaginary
$$\frac { { z }_{ 2 }-{ 2z }_{ 2 } }{ { z }_{ 2 }-{ z }_{ 1 }{ z }_{ 2 } } $$ is unimodular then
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$$|{ z }_{ 2 }|=2$$
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$$|{ z }_{ 1 }|=1$$
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Both A and B
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None of these
If z is a complex number of unit modules and argument $$\theta $$, then the real part of $$\dfrac { z(1-\bar { z } ) }{ z(1+z) } $$ is :
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$${ -2sin }^{ 2 }\dfrac { \theta }{ 2 } $$
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$${ 2sin }^{ 2 }\dfrac { \theta }{ 2 } $$
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$$1+cos\dfrac { \theta }{ 2 } $$
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$$1-cos\dfrac { \theta }{ 2 } $$
This equation $$(x-5)^{11}+(x-5^{2})^{11}+....+(x-5^{11})^{11}=0$$ has
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all the roots real
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one real and 10 imaginary roots
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real roots namely $$x=5,5^{2}....,5^{9},5^{10},5^{11}$$
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none
If z= $$\dfrac { 1 }{ { \left( 2+3i \right) }^{ 2 } } ,\quad then\left| z \right| $$=
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$$\dfrac { 1 }{ 13 } $$
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$$\dfrac { 1 }{ 5 } $$
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$$\dfrac { 1 }{ 12 } $$
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none of these
The complex number z satisfies the equation z + |z| = 2 + 8i. Then the value of |z| is
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15
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16
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17
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18
Explanation
$$z+\left| z \right| =2+8i\quad \longrightarrow \left( 1 \right) $$
let $$z=x+iy\quad \longrightarrow \left( 2 \right) $$
then, $$\left| z \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \quad \longrightarrow \left( 3 \right) $$
$$\therefore$$ Substituting $$(2)$$ & $$(3)$$ in $$(1)$$ we get
$$x+iy+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =2+8i$$
$$\Rightarrow x+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } +iy=2+8i$$
Comparing $$LHS$$ & $$RHS$$ we get
$$y=8\longrightarrow \left( 4 \right) $$ and $$x+\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =2\quad \longrightarrow \left( 5 \right) $$
Substituting the value of $$y$$ in the equation $$(5)$$
We get
$$x+\sqrt { { x }^{ 2 }+64 } =2$$
$$\sqrt { { x }^{ 2 }+64 } =2-x$$
Squaring both sides,
$${ x }^{ 2 }+64=4+{ x }^{ 2 }-4x$$
$$\Rightarrow 4x=-60$$
$$\Rightarrow x=-15$$
$$\therefore$$ $$z=-15+8i$$
$$\left| z \right| =\sqrt { 225+164 } =\sqrt { 289 } =17$$ units.
The complex number $$z$$ satisfies $$z+|z|=2+8i$$. The value of $$|z|$$ is
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10
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13
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17
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23
The modulus of the complex number z such that $$\left | z+3-i \right |=1 $$ and arg $$z=\pi $$ is equal to
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1
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2
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9
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4
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3
For two unimodular complex numbers $$ z_1 $$ and $$ z_2 $$,
$$ \left[ \begin{matrix} \bar { z_{ 1 } } & -z_{ 2 } \\ \bar { z_{ 1 } } & z_{ 1 } \end{matrix} \right] ^{ -1 } \left[ \begin{matrix} z_1 & z_2 \\- \bar { z_{ 2 } } & \bar{ z_1} \end{matrix} \right] ^{ -1 } $$ is equal to
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$$ \begin{bmatrix} z_1 & z_2 \\ \bar {z_1} & \bar {z_2} \end{bmatrix} $$
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$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
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$$ \begin{bmatrix} \dfrac{1}{2} & 0 \\ 0 & \dfrac{1}{2} \end{bmatrix} $$
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None of these
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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