QBManager

Complex Numbers and Quadratic Equation

117966 If $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$ , then $amp (z_{1} + z_{2}) =$

1

\frac{π}{12}

2

\frac{π}{15}

3

\frac{π}{6}

4

\frac{π}{4}

Explanation:

A We have, $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$
So, $z_{1} + z_{2} = (\sqrt{3} + 1) + i (\sqrt{3} - 1)$
Putting $r \cos θ = \sqrt{3} + 1$ and $r \sin θ = \sqrt{3} - 1$
$∴ \tan θ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$
$= \frac{(\sqrt{3} - 1)^{2}}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2}$
$\tan θ = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan θ = \tan \frac{π}{12}$
$Amplitude (θ) = \frac{π}{12}$

COMEDK-2011

Complex Numbers and Quadratic Equation

117958 If $z (2 - i) = 3 + i, z^{20} =$

1

1 - i

2 -1024

3 1024

4

1 + i

Explanation:

B Given that,
$z (2 - i) = (3 + i)$
$z = \frac{3 + i}{2 - i}$
Now, multiply and divide by $(2 + i)$
$z = \frac{(3 + i)}{(2 - i)} \times \frac{(2 + i)}{(2 + i)}$
$= \frac{6 + 5 i - i^{2}}{4 - i^{2}} = \frac{5 + 5 i}{5} = 1 + i$
$z = 1 + i$
Squaring both side
$z^{2} = (1 + i)^{2}$
$= 1 + i^{2} + 2 i = 2 i$
$z^{2} = 2 i$
Value of $z^{20}$
$z^{20} = {(z^{2})}^{10}$
$= (2 i)^{10}$
$= 2^{10} i^{10}$
$z^{20} = - 1024$
$(i^{2} = - 1)$

Karnataka CET-2002

Complex Numbers and Quadratic Equation

117959 If $z = r^{i θ}$ then $| e^{i z} |$ is equal to

1

e^{- rsin θ}

2

{re}^{- r \sin θ}

3

e^{- r \cos θ}

4

{re}^{- r \cos θ}

Explanation:

A Given, $z = {re}^{i θ}$
$z = r (\cos θ + i \sin θ)$
$iz = ir (\cos θ + i \sin θ)$
$iz = r (i \cos θ + i^{2} \sin θ)$
$iz = r (i \cos θ - \sin θ)$
$iz = i r \cos θ - r \sin θ$
$| e^{i} z | = | e^{- r \sin θ + ir \cos θ} | = | e^{- r \sin θ} | | e^{i \cos θ} |$
$= e^{- r \sin θ} \sqrt{\cos^{2} (r \cos θ) + \sin^{2} (r \cos θ)}$
$= e^{- r \sin θ}$ So, $| e^{iz} |$ is equal to $e^{- r \sin θ}$ .

Kerala CEE - 2005]

Complex Numbers and Quadratic Equation

117960 The region of the argand diagram defined by $| z - 1 | + | z + 1 | \leq 4$ is

1 Interior of an ellipse

2 exterior of a circle

3 interior and bondry of an ellipse

4 interior of a circle

Explanation:

C Let, $z = x +$ iy
Given, $| z - 1 | + | z + 1 | \leq 4$
$∣ x + iy - 1 | + | x + i y + 1 ∣\leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x + 1)^{2} + y^{2}} \leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} \leq 4 - \sqrt{(x + 1)^{2} + y^{2}}$
Squaring on both side,
$(x - 1)^{2} + y^{2} \leq {(4 - \sqrt{(x + 1)^{2} + y^{2}})}^{2}$
$x^{2} + 1 - 2 x + y^{2} \leq 16 + x^{2} + 1 + 2 x + y^{2}$
$- 8 \sqrt{(x + 1)^{2} + y^{2}}$
$8 \sqrt{(x + 1)^{2} + y^{2}} \leq 4 x + 16$
$4 [(x + 1)^{2} + y^{2}] \leq (x + 4)^{2}$
$4 x^{2} + 4 + 8 x + 4 y^{2} \leq x^{2} + 16 + 8 x$
$3 x^{2} + 4 y^{2} \leq 12$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} \leq 1$
Which is an ellipse.
So, the region is on the bondry and interior of ellipse.

SRM JEEE-2011

Complex Numbers and Quadratic Equation

117961 The real part of $(1 - \cos θ + i \sin θ)^{- 1}$ is

1

\frac{1}{2}

2

\frac{1}{1 + \cos θ}

3

\tan \frac{θ}{2}

4

\cot \frac{θ}{2}

Explanation:

A Given,
$(1 - \cos θ + i \sin θ)^{- 1}$
$= \frac{1}{(1 - \cos θ + i \sin θ)}$
$= \frac{1}{(1 - \cos θ) + i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} + i \cdot 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{1}{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2}) (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} [\sin \frac{θ}{2} - i \cos \frac{θ}{2}]$
$= \frac{1}{2} - \frac{1}{2} i \cot \frac{θ}{2}$ So, real part of $(1 - \cos θ + i \sin θ)^{- 1} = \frac{1}{2}$

Karnataka CET-2016

Complex Numbers and Quadratic Equation

117966 If $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$ , then $amp (z_{1} + z_{2}) =$

1

\frac{π}{12}

2

\frac{π}{15}

3

\frac{π}{6}

4

\frac{π}{4}

Explanation:

A We have, $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$
So, $z_{1} + z_{2} = (\sqrt{3} + 1) + i (\sqrt{3} - 1)$
Putting $r \cos θ = \sqrt{3} + 1$ and $r \sin θ = \sqrt{3} - 1$
$∴ \tan θ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$
$= \frac{(\sqrt{3} - 1)^{2}}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2}$
$\tan θ = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan θ = \tan \frac{π}{12}$
$Amplitude (θ) = \frac{π}{12}$

COMEDK-2011

Complex Numbers and Quadratic Equation

117958 If $z (2 - i) = 3 + i, z^{20} =$

1

1 - i

2 -1024

3 1024

4

1 + i

Explanation:

B Given that,
$z (2 - i) = (3 + i)$
$z = \frac{3 + i}{2 - i}$
Now, multiply and divide by $(2 + i)$
$z = \frac{(3 + i)}{(2 - i)} \times \frac{(2 + i)}{(2 + i)}$
$= \frac{6 + 5 i - i^{2}}{4 - i^{2}} = \frac{5 + 5 i}{5} = 1 + i$
$z = 1 + i$
Squaring both side
$z^{2} = (1 + i)^{2}$
$= 1 + i^{2} + 2 i = 2 i$
$z^{2} = 2 i$
Value of $z^{20}$
$z^{20} = {(z^{2})}^{10}$
$= (2 i)^{10}$
$= 2^{10} i^{10}$
$z^{20} = - 1024$
$(i^{2} = - 1)$

Karnataka CET-2002

Complex Numbers and Quadratic Equation

117959 If $z = r^{i θ}$ then $| e^{i z} |$ is equal to

1

e^{- rsin θ}

2

{re}^{- r \sin θ}

3

e^{- r \cos θ}

4

{re}^{- r \cos θ}

Explanation:

A Given, $z = {re}^{i θ}$
$z = r (\cos θ + i \sin θ)$
$iz = ir (\cos θ + i \sin θ)$
$iz = r (i \cos θ + i^{2} \sin θ)$
$iz = r (i \cos θ - \sin θ)$
$iz = i r \cos θ - r \sin θ$
$| e^{i} z | = | e^{- r \sin θ + ir \cos θ} | = | e^{- r \sin θ} | | e^{i \cos θ} |$
$= e^{- r \sin θ} \sqrt{\cos^{2} (r \cos θ) + \sin^{2} (r \cos θ)}$
$= e^{- r \sin θ}$ So, $| e^{iz} |$ is equal to $e^{- r \sin θ}$ .

Kerala CEE - 2005]

Complex Numbers and Quadratic Equation

117960 The region of the argand diagram defined by $| z - 1 | + | z + 1 | \leq 4$ is

1 Interior of an ellipse

2 exterior of a circle

3 interior and bondry of an ellipse

4 interior of a circle

Explanation:

C Let, $z = x +$ iy
Given, $| z - 1 | + | z + 1 | \leq 4$
$∣ x + iy - 1 | + | x + i y + 1 ∣\leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x + 1)^{2} + y^{2}} \leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} \leq 4 - \sqrt{(x + 1)^{2} + y^{2}}$
Squaring on both side,
$(x - 1)^{2} + y^{2} \leq {(4 - \sqrt{(x + 1)^{2} + y^{2}})}^{2}$
$x^{2} + 1 - 2 x + y^{2} \leq 16 + x^{2} + 1 + 2 x + y^{2}$
$- 8 \sqrt{(x + 1)^{2} + y^{2}}$
$8 \sqrt{(x + 1)^{2} + y^{2}} \leq 4 x + 16$
$4 [(x + 1)^{2} + y^{2}] \leq (x + 4)^{2}$
$4 x^{2} + 4 + 8 x + 4 y^{2} \leq x^{2} + 16 + 8 x$
$3 x^{2} + 4 y^{2} \leq 12$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} \leq 1$
Which is an ellipse.
So, the region is on the bondry and interior of ellipse.

SRM JEEE-2011

Complex Numbers and Quadratic Equation

117961 The real part of $(1 - \cos θ + i \sin θ)^{- 1}$ is

1

\frac{1}{2}

2

\frac{1}{1 + \cos θ}

3

\tan \frac{θ}{2}

4

\cot \frac{θ}{2}

Explanation:

A Given,
$(1 - \cos θ + i \sin θ)^{- 1}$
$= \frac{1}{(1 - \cos θ + i \sin θ)}$
$= \frac{1}{(1 - \cos θ) + i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} + i \cdot 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{1}{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2}) (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} [\sin \frac{θ}{2} - i \cos \frac{θ}{2}]$
$= \frac{1}{2} - \frac{1}{2} i \cot \frac{θ}{2}$ So, real part of $(1 - \cos θ + i \sin θ)^{- 1} = \frac{1}{2}$

Karnataka CET-2016

Complex Numbers and Quadratic Equation

117966 If $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$ , then $amp (z_{1} + z_{2}) =$

1

\frac{π}{12}

2

\frac{π}{15}

3

\frac{π}{6}

4

\frac{π}{4}

Explanation:

A We have, $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$
So, $z_{1} + z_{2} = (\sqrt{3} + 1) + i (\sqrt{3} - 1)$
Putting $r \cos θ = \sqrt{3} + 1$ and $r \sin θ = \sqrt{3} - 1$
$∴ \tan θ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$
$= \frac{(\sqrt{3} - 1)^{2}}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2}$
$\tan θ = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan θ = \tan \frac{π}{12}$
$Amplitude (θ) = \frac{π}{12}$

COMEDK-2011

Complex Numbers and Quadratic Equation

117958 If $z (2 - i) = 3 + i, z^{20} =$

1

1 - i

2 -1024

3 1024

4

1 + i

Explanation:

B Given that,
$z (2 - i) = (3 + i)$
$z = \frac{3 + i}{2 - i}$
Now, multiply and divide by $(2 + i)$
$z = \frac{(3 + i)}{(2 - i)} \times \frac{(2 + i)}{(2 + i)}$
$= \frac{6 + 5 i - i^{2}}{4 - i^{2}} = \frac{5 + 5 i}{5} = 1 + i$
$z = 1 + i$
Squaring both side
$z^{2} = (1 + i)^{2}$
$= 1 + i^{2} + 2 i = 2 i$
$z^{2} = 2 i$
Value of $z^{20}$
$z^{20} = {(z^{2})}^{10}$
$= (2 i)^{10}$
$= 2^{10} i^{10}$
$z^{20} = - 1024$
$(i^{2} = - 1)$

Karnataka CET-2002

Complex Numbers and Quadratic Equation

117959 If $z = r^{i θ}$ then $| e^{i z} |$ is equal to

1

e^{- rsin θ}

2

{re}^{- r \sin θ}

3

e^{- r \cos θ}

4

{re}^{- r \cos θ}

Explanation:

A Given, $z = {re}^{i θ}$
$z = r (\cos θ + i \sin θ)$
$iz = ir (\cos θ + i \sin θ)$
$iz = r (i \cos θ + i^{2} \sin θ)$
$iz = r (i \cos θ - \sin θ)$
$iz = i r \cos θ - r \sin θ$
$| e^{i} z | = | e^{- r \sin θ + ir \cos θ} | = | e^{- r \sin θ} | | e^{i \cos θ} |$
$= e^{- r \sin θ} \sqrt{\cos^{2} (r \cos θ) + \sin^{2} (r \cos θ)}$
$= e^{- r \sin θ}$ So, $| e^{iz} |$ is equal to $e^{- r \sin θ}$ .

Kerala CEE - 2005]

Complex Numbers and Quadratic Equation

117960 The region of the argand diagram defined by $| z - 1 | + | z + 1 | \leq 4$ is

1 Interior of an ellipse

2 exterior of a circle

3 interior and bondry of an ellipse

4 interior of a circle

Explanation:

C Let, $z = x +$ iy
Given, $| z - 1 | + | z + 1 | \leq 4$
$∣ x + iy - 1 | + | x + i y + 1 ∣\leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x + 1)^{2} + y^{2}} \leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} \leq 4 - \sqrt{(x + 1)^{2} + y^{2}}$
Squaring on both side,
$(x - 1)^{2} + y^{2} \leq {(4 - \sqrt{(x + 1)^{2} + y^{2}})}^{2}$
$x^{2} + 1 - 2 x + y^{2} \leq 16 + x^{2} + 1 + 2 x + y^{2}$
$- 8 \sqrt{(x + 1)^{2} + y^{2}}$
$8 \sqrt{(x + 1)^{2} + y^{2}} \leq 4 x + 16$
$4 [(x + 1)^{2} + y^{2}] \leq (x + 4)^{2}$
$4 x^{2} + 4 + 8 x + 4 y^{2} \leq x^{2} + 16 + 8 x$
$3 x^{2} + 4 y^{2} \leq 12$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} \leq 1$
Which is an ellipse.
So, the region is on the bondry and interior of ellipse.

SRM JEEE-2011

Complex Numbers and Quadratic Equation

117961 The real part of $(1 - \cos θ + i \sin θ)^{- 1}$ is

1

\frac{1}{2}

2

\frac{1}{1 + \cos θ}

3

\tan \frac{θ}{2}

4

\cot \frac{θ}{2}

Explanation:

A Given,
$(1 - \cos θ + i \sin θ)^{- 1}$
$= \frac{1}{(1 - \cos θ + i \sin θ)}$
$= \frac{1}{(1 - \cos θ) + i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} + i \cdot 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{1}{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2}) (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} [\sin \frac{θ}{2} - i \cos \frac{θ}{2}]$
$= \frac{1}{2} - \frac{1}{2} i \cot \frac{θ}{2}$ So, real part of $(1 - \cos θ + i \sin θ)^{- 1} = \frac{1}{2}$

Karnataka CET-2016

Complex Numbers and Quadratic Equation

117966 If $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$ , then $amp (z_{1} + z_{2}) =$

1

\frac{π}{12}

2

\frac{π}{15}

3

\frac{π}{6}

4

\frac{π}{4}

Explanation:

A We have, $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$
So, $z_{1} + z_{2} = (\sqrt{3} + 1) + i (\sqrt{3} - 1)$
Putting $r \cos θ = \sqrt{3} + 1$ and $r \sin θ = \sqrt{3} - 1$
$∴ \tan θ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$
$= \frac{(\sqrt{3} - 1)^{2}}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2}$
$\tan θ = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan θ = \tan \frac{π}{12}$
$Amplitude (θ) = \frac{π}{12}$

COMEDK-2011

Complex Numbers and Quadratic Equation

117958 If $z (2 - i) = 3 + i, z^{20} =$

1

1 - i

2 -1024

3 1024

4

1 + i

Explanation:

B Given that,
$z (2 - i) = (3 + i)$
$z = \frac{3 + i}{2 - i}$
Now, multiply and divide by $(2 + i)$
$z = \frac{(3 + i)}{(2 - i)} \times \frac{(2 + i)}{(2 + i)}$
$= \frac{6 + 5 i - i^{2}}{4 - i^{2}} = \frac{5 + 5 i}{5} = 1 + i$
$z = 1 + i$
Squaring both side
$z^{2} = (1 + i)^{2}$
$= 1 + i^{2} + 2 i = 2 i$
$z^{2} = 2 i$
Value of $z^{20}$
$z^{20} = {(z^{2})}^{10}$
$= (2 i)^{10}$
$= 2^{10} i^{10}$
$z^{20} = - 1024$
$(i^{2} = - 1)$

Karnataka CET-2002

Complex Numbers and Quadratic Equation

117959 If $z = r^{i θ}$ then $| e^{i z} |$ is equal to

1

e^{- rsin θ}

2

{re}^{- r \sin θ}

3

e^{- r \cos θ}

4

{re}^{- r \cos θ}

Explanation:

A Given, $z = {re}^{i θ}$
$z = r (\cos θ + i \sin θ)$
$iz = ir (\cos θ + i \sin θ)$
$iz = r (i \cos θ + i^{2} \sin θ)$
$iz = r (i \cos θ - \sin θ)$
$iz = i r \cos θ - r \sin θ$
$| e^{i} z | = | e^{- r \sin θ + ir \cos θ} | = | e^{- r \sin θ} | | e^{i \cos θ} |$
$= e^{- r \sin θ} \sqrt{\cos^{2} (r \cos θ) + \sin^{2} (r \cos θ)}$
$= e^{- r \sin θ}$ So, $| e^{iz} |$ is equal to $e^{- r \sin θ}$ .

Kerala CEE - 2005]

Complex Numbers and Quadratic Equation

117960 The region of the argand diagram defined by $| z - 1 | + | z + 1 | \leq 4$ is

1 Interior of an ellipse

2 exterior of a circle

3 interior and bondry of an ellipse

4 interior of a circle

Explanation:

C Let, $z = x +$ iy
Given, $| z - 1 | + | z + 1 | \leq 4$
$∣ x + iy - 1 | + | x + i y + 1 ∣\leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x + 1)^{2} + y^{2}} \leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} \leq 4 - \sqrt{(x + 1)^{2} + y^{2}}$
Squaring on both side,
$(x - 1)^{2} + y^{2} \leq {(4 - \sqrt{(x + 1)^{2} + y^{2}})}^{2}$
$x^{2} + 1 - 2 x + y^{2} \leq 16 + x^{2} + 1 + 2 x + y^{2}$
$- 8 \sqrt{(x + 1)^{2} + y^{2}}$
$8 \sqrt{(x + 1)^{2} + y^{2}} \leq 4 x + 16$
$4 [(x + 1)^{2} + y^{2}] \leq (x + 4)^{2}$
$4 x^{2} + 4 + 8 x + 4 y^{2} \leq x^{2} + 16 + 8 x$
$3 x^{2} + 4 y^{2} \leq 12$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} \leq 1$
Which is an ellipse.
So, the region is on the bondry and interior of ellipse.

SRM JEEE-2011

Complex Numbers and Quadratic Equation

117961 The real part of $(1 - \cos θ + i \sin θ)^{- 1}$ is

1

\frac{1}{2}

2

\frac{1}{1 + \cos θ}

3

\tan \frac{θ}{2}

4

\cot \frac{θ}{2}

Explanation:

A Given,
$(1 - \cos θ + i \sin θ)^{- 1}$
$= \frac{1}{(1 - \cos θ + i \sin θ)}$
$= \frac{1}{(1 - \cos θ) + i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} + i \cdot 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{1}{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2}) (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} [\sin \frac{θ}{2} - i \cos \frac{θ}{2}]$
$= \frac{1}{2} - \frac{1}{2} i \cot \frac{θ}{2}$ So, real part of $(1 - \cos θ + i \sin θ)^{- 1} = \frac{1}{2}$

Karnataka CET-2016

Complex Numbers and Quadratic Equation

117966 If $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$ , then $amp (z_{1} + z_{2}) =$

1

\frac{π}{12}

2

\frac{π}{15}

3

\frac{π}{6}

4

\frac{π}{4}

Explanation:

A We have, $z_{1} = \sqrt{3} - i$ and $z_{2} = 1 + i \sqrt{3}$
So, $z_{1} + z_{2} = (\sqrt{3} + 1) + i (\sqrt{3} - 1)$
Putting $r \cos θ = \sqrt{3} + 1$ and $r \sin θ = \sqrt{3} - 1$
$∴ \tan θ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$
$= \frac{(\sqrt{3} - 1)^{2}}{3 - 1} = \frac{3 + 1 - 2 \sqrt{3}}{2}$
$\tan θ = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$
$\tan θ = \tan \frac{π}{12}$
$Amplitude (θ) = \frac{π}{12}$

COMEDK-2011

Complex Numbers and Quadratic Equation

117958 If $z (2 - i) = 3 + i, z^{20} =$

1

1 - i

2 -1024

3 1024

4

1 + i

Explanation:

B Given that,
$z (2 - i) = (3 + i)$
$z = \frac{3 + i}{2 - i}$
Now, multiply and divide by $(2 + i)$
$z = \frac{(3 + i)}{(2 - i)} \times \frac{(2 + i)}{(2 + i)}$
$= \frac{6 + 5 i - i^{2}}{4 - i^{2}} = \frac{5 + 5 i}{5} = 1 + i$
$z = 1 + i$
Squaring both side
$z^{2} = (1 + i)^{2}$
$= 1 + i^{2} + 2 i = 2 i$
$z^{2} = 2 i$
Value of $z^{20}$
$z^{20} = {(z^{2})}^{10}$
$= (2 i)^{10}$
$= 2^{10} i^{10}$
$z^{20} = - 1024$
$(i^{2} = - 1)$

Karnataka CET-2002

Complex Numbers and Quadratic Equation

117959 If $z = r^{i θ}$ then $| e^{i z} |$ is equal to

1

e^{- rsin θ}

2

{re}^{- r \sin θ}

3

e^{- r \cos θ}

4

{re}^{- r \cos θ}

Explanation:

A Given, $z = {re}^{i θ}$
$z = r (\cos θ + i \sin θ)$
$iz = ir (\cos θ + i \sin θ)$
$iz = r (i \cos θ + i^{2} \sin θ)$
$iz = r (i \cos θ - \sin θ)$
$iz = i r \cos θ - r \sin θ$
$| e^{i} z | = | e^{- r \sin θ + ir \cos θ} | = | e^{- r \sin θ} | | e^{i \cos θ} |$
$= e^{- r \sin θ} \sqrt{\cos^{2} (r \cos θ) + \sin^{2} (r \cos θ)}$
$= e^{- r \sin θ}$ So, $| e^{iz} |$ is equal to $e^{- r \sin θ}$ .

Kerala CEE - 2005]

Complex Numbers and Quadratic Equation

117960 The region of the argand diagram defined by $| z - 1 | + | z + 1 | \leq 4$ is

1 Interior of an ellipse

2 exterior of a circle

3 interior and bondry of an ellipse

4 interior of a circle

Explanation:

C Let, $z = x +$ iy
Given, $| z - 1 | + | z + 1 | \leq 4$
$∣ x + iy - 1 | + | x + i y + 1 ∣\leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} + \sqrt{(x + 1)^{2} + y^{2}} \leq 4$
$\sqrt{(x - 1)^{2} + y^{2}} \leq 4 - \sqrt{(x + 1)^{2} + y^{2}}$
Squaring on both side,
$(x - 1)^{2} + y^{2} \leq {(4 - \sqrt{(x + 1)^{2} + y^{2}})}^{2}$
$x^{2} + 1 - 2 x + y^{2} \leq 16 + x^{2} + 1 + 2 x + y^{2}$
$- 8 \sqrt{(x + 1)^{2} + y^{2}}$
$8 \sqrt{(x + 1)^{2} + y^{2}} \leq 4 x + 16$
$4 [(x + 1)^{2} + y^{2}] \leq (x + 4)^{2}$
$4 x^{2} + 4 + 8 x + 4 y^{2} \leq x^{2} + 16 + 8 x$
$3 x^{2} + 4 y^{2} \leq 12$
$\frac{x^{2}}{4} + \frac{y^{2}}{3} \leq 1$
Which is an ellipse.
So, the region is on the bondry and interior of ellipse.

SRM JEEE-2011

Complex Numbers and Quadratic Equation

117961 The real part of $(1 - \cos θ + i \sin θ)^{- 1}$ is

1

\frac{1}{2}

2

\frac{1}{1 + \cos θ}

3

\tan \frac{θ}{2}

4

\cot \frac{θ}{2}

Explanation:

A Given,
$(1 - \cos θ + i \sin θ)^{- 1}$
$= \frac{1}{(1 - \cos θ + i \sin θ)}$
$= \frac{1}{(1 - \cos θ) + i \sin θ} = \frac{1}{2 \sin^{2} \frac{θ}{2} + i \cdot 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{1}{\sin \frac{θ}{2} + i \cos \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{(\sin \frac{θ}{2} + i \cos \frac{θ}{2}) (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}]$
$= \frac{1}{2 \sin \frac{θ}{2}} \cdot [\frac{\sin \frac{θ}{2} - i \cos \frac{θ}{2}}{\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2}}]$
$= \frac{1}{2 \sin \frac{θ}{2}} [\sin \frac{θ}{2} - i \cos \frac{θ}{2}]$
$= \frac{1}{2} - \frac{1}{2} i \cot \frac{θ}{2}$ So, real part of $(1 - \cos θ + i \sin θ)^{- 1} = \frac{1}{2}$

Karnataka CET-2016