(a) The sum and product of the given complex numbers.

Question

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Answer 1

Question

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

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Answer 2

Question

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

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Answer 3

Question

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

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Answer 4

Question

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

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Answer 5

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

Answer 6

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

Answer 7

Chapter 9.1, Problem 1PS

To determine

(a)

The sum and product of the given complex numbers.

Answer 8

Expert Solution

Answer to Problem 1PS

z1 + z2 =4 and z1z2 =5

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

z1=2+i,z2=2−i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=2+i,z2=2−i

z1 + z2 =(2+i)+(2−i)⇒z1 + z2 =(2+2)+i(1−1)⇒z1 + z2 =4+i(0)∴z1 + z2 =4

z1 z2 =(2+i)(2−i)⇒z1z2 ={2×2 − 1(−1)}+i{2(−1)+2(1)}⇒z1z2 ={4+1}+i{−2+2}⇒z1z2 =5+i{0}∴z1z2 =5

Answer 13

To determine

(b)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

Answer 14

To determine

(b)

The sum and product of the given complex numbers.

Answer 15

Expert Solution

Answer to Problem 1PS

z1 + z2 =−2+2i and z1z2 =−2i

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

z1=−1+i,z2=−1+i

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=−1+i,z2=−1+i

z1 + z2 =(−1+i)+(−1+i)⇒z1 + z2 =(−1−1)+i(1+1)⇒z1 + z2 =−2+i(2)∴z1 + z2 =−2+2i

z1 z2 =(−1+i)(−1+i)⇒z1z2 ={(−1)×(−1) − 1×1}+i{(−1)×1+ (−1)×1}⇒z1z2 ={1−1}+i{−1−1}⇒z1z2 =0+i{−2}∴z1z2 =−2i

Answer 20

To determine

(c)

The sum and product of the given complex numbers.

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1

Answer 21

To determine

(c)

The sum and product of the given complex numbers.

Answer 22

Expert Solution

Answer to Problem 1PS

z1 + z2 =2cosθ and z1z2 =1

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

z1=cosθ+isinθ,z2=cosθ−isinθ

Concept Used:

If z1=x1+iy1,z2=x2+iy2 are any two complex numbers, then

z1 + z2 =(x1+iy1)+(x2+iy2)⇒z1 + z2 =(x1+x2)+(iy1+iy2)⇒z1 + z2 =(x1+x2)+i(y1+y2)

z1z2 =(x1+iy1)(x2+iy2)⇒z1z2 =x1(x2+iy2)+iy1(x2+iy2)⇒z1z2 =x1x2+ix1y2+ix2y1+i2y1y2⇒z1z2 =x1x2+ix1y2+ix2y1−y1y2 [∵ i2=−1]⇒z1z2 =(x1x2 − y1y2)+i(x1y2+x2y1)

Calculation:

Here, we have

z1=cosθ+isinθ,z2=cosθ−isinθ

z1 + z2 =(cosθ+isinθ)+ (cosθ−isinθ)⇒z1 + z2 =(cosθ+cosθ)+i(sinθ−sinθ)⇒z1 + z2 =2cosθ+i(0)∴z1 + z2 =2cosθ

z1 z2 =(cosθ+isinθ) (cosθ−isinθ)⇒z1z2 =(cosθ)2−(isinθ)2 [∵(a+b)(a−b)=a2−b2]⇒z1z2 =cos2θ−i2sin2θ⇒z1z2 =cos2θ−(−1)sin2θ [∵i2=−1]⇒z1z2 =cos2θ+sin2θ [∵cos2θ+sin2θ =1]∴z1z2 =1