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1

WB JEE 2009

MCQ (Single Correct Answer)

The modulus of $${{1 - i} \over {3 + i}} + {{4i} \over 5}$$ is

A
$$\sqrt 5 $$ unit
B
$${{\sqrt {11} } \over 5}$$ unit
C
$${{\sqrt 5 } \over 5}$$ unit
D
$${{\sqrt {12} } \over 5}$$ unit

Explanation

$$z = {{1 - i} \over {3 + i}} + {{4i} \over 5} = {1 \over 5} + {2 \over 5}i$$

$$|z| = \sqrt {{1 \over {25}} + {4 \over {25}}} = {1 \over {\sqrt 5 }} = {{\sqrt 5 } \over 5}$$ units.

2

WB JEE 2009

MCQ (Single Correct Answer)

If $$i = \sqrt { - 1} $$ and n is positive integer, then $${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}$$ is equal to

A
1
B
i
C
in
D
0

Explanation

$${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}$$

$$ = {i^n}(1 + i + {i^2} + {i^3}) = {i^n}(1 + i - 1 - i) = 0$$

3

WB JEE 2008

MCQ (Single Correct Answer)

If 1, $$\omega$$, $$\omega$$2 are cube roots of unity, then $$\left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr } } \right|$$ has value

A
0
B
$$\omega$$
C
$$\omega$$2
D
$$\omega$$ + $$\omega$$2

Explanation

Property of $$\omega$$ is $$\omega$$3 = 1 and 1 + $$\omega$$ + $$\omega$$2 = 0

$$\therefore$$ $$\left| {\matrix{ 1 & {{\omega ^n}} & {{\omega ^{2n}}} \cr {{\omega ^{2n}}} & 1 & {{\omega ^n}} \cr {{\omega ^n}} & {{\omega ^{2n}}} & 1 \cr } } \right|$$

$$ = 1(1 - {\omega ^n} \times {\omega ^{2n}}) - {\omega ^n}({\omega ^{2n}} - {\omega ^n} \times {\omega ^n}) + {\omega ^{2n}}({\omega ^{2n}} \times {\omega ^{2n}} - {\omega ^n})$$

$$ = 1 - {\omega ^{3n}} - {\omega ^{3n}} + {\omega ^{3n}} + {\omega ^{6n}} - {\omega ^{3n}} = 1 - 2{({\omega ^3})^n} + {({\omega ^3})^{2n}}$$

$$ = 1 - 2{(1)^n} + {(1)^{2n}} = 1 - 2 + 1 = 0$$.

4

WB JEE 2008

MCQ (Single Correct Answer)

For two complex numbers z1, z2 the relation $$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$ holds if

A
$$\arg ({z_1}) = \arg ({z_2})$$
B
$$\arg ({z_1}) + \arg ({z_2}) = {\pi \over 2}$$
C
$${z_1}{z_2} = 1$$
D
$$\left| {{z_1}} \right| = \left| {{z_2}} \right|$$

Explanation

Let $${z_1} = {r_1}(\cos {\theta _1} + i\sin {\theta _1})$$ and $${z_2} = {r_2}(\cos {\theta _2} + i\sin {\theta _2})$$

$$\therefore$$ $${z_1} + {z_2} = ({r_1}\cos {\theta _1} + {r_2}\cos {\theta _2}) + i({r_1}\sin {\theta _1} + {r_2}\sin {\theta _2})$$

$$\left| {{z_1} + {z_2}} \right| = \sqrt {{{({r_1}\cos {\theta _1} + {r_2}\cos {\theta _2})}^2} + {{({r_1}\sin {\theta _1} + {r_2}\sin {\theta _2})}^2}} $$

$$ = \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2})} $$

$$\left| {{z_1}} \right| = {r_1},\,\left| {{z_2}} \right| = {r_2}$$

$$ \Rightarrow \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2})} = {r_1} + {r_2}$$ (given $$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$$)

(squaring both sides)

$$ \Rightarrow r_1^2 + r_2^2 + 2{r_1}{r_2}\cos ({\theta _1} - {\theta _2}) = r_1^2 + r_2^2 + 2{r_1}{r_2}$$

$$ \Rightarrow \cos ({\theta _1} - {\theta _2}) = \cos 0^\circ \Rightarrow {\theta _1} = {\theta _2}$$

$$ \Rightarrow \arg ({z_1}) = \arg ({z_2})$$.

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