# To prove: The property z + w ¯ = z ¯ + w ¯ . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter I, Problem 18E

(a)

To determine

## To prove: The property z+w¯=z¯+w¯.

Expert Solution

### Explanation of Solution

Let z=a+bi and w=c+di.

Consider the left hand of the property and simplified as follows.

z+w¯=(a+bi)+(c+di)¯=(a+c)+(b+d)i¯=(a+c)(b+d)i=(abi)+(cdi)

That is, z+w¯=(abi)+(cdi) (1)

Consider the right hand of the property and simplified as follows.

z¯+w¯=(a+bi)¯+(c+di)¯=(abi)+(cdi)

That is, z¯+w¯=(abi)+(cdi) (2)

From equation (1) and (2) it is observed that z+w¯=z¯+w¯.

Hence the proof.

(b)

To determine

Expert Solution

### Explanation of Solution

Let z=a+bi and w=c+di.

Consider the left hand of the property and simplified as follows.

Consider the right hand of the property and simplified as follows.

From equation (1) and (2) it is observed that zw¯=z¯w¯.

Hence the proof.

(c)

To determine

Expert Solution

### Explanation of Solution

Let Sn be the statement that (zn)¯=(z¯)n.

To check the statement is true for n=1.

(z1)¯=(z¯)1

That is,  the statement Sn is true for n=1.

Assume that the statement Sn is true for n=k.

That is, (zk)¯=(z¯)k.

To show the statement is true for n=k+1.

(zk+1)¯=(zkz)¯=(zk)¯(z¯)=(z¯)k(z¯)[(zk)¯=(z¯)k]=(z¯)k+1

Thus, the statement is true for n=k+1.

Therefore, the statement Sn is true for all values of n by mathematical induction.

Hence the formula is proved.

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