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1

WB JEE 2009

MCQ (Single Correct Answer)

For any complex number z, the minimum value of $$|z| + |z - 1|$$ is

A
0
B
1
C
2
D
$$-$$1

Explanation

$$\because$$ $$||{z_1}| - |{z_2}|| \le |{z_1} - {z_2}|$$

$$ \Rightarrow ||z| - 1| \le |z - 1|$$

$$ \Rightarrow - |z - 1| \le |z| - 1| \le |z - 1|$$

$$ \Rightarrow - |z - 1| \le |z| - 1 \Rightarrow 1 \le |z| + |z - 1|$$

So minimum value of $$|z| + |z - 1|$$ is 1

2

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.

3

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

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

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

4

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

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

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

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

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

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