BuyFind

Single Variable Calculus: Concepts...

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

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

To prove: The property zw¯=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.

zw¯=(a+bi)(c+di)¯=(acbd)+(ad+bc)i¯=(acbd)+(ad+bc)i

That is, zw¯=(acbd)+(ad+bc)i (1)

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

z¯w¯=(a+bi)¯(c+di)¯=(abi)(cdi)=(acbd)+(ad+bc)i

That is, z¯w¯=(acbd)+(ad+bc)i (2)

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

Hence the proof.

(c)

To determine

To prove: The property (zn)¯=(z¯)n.

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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