Exercise 76 Page 393

Practice makes perfect

We want to calculate the value of the given numeric expression using the imaginary number i. To do so, we will first recall that the imaginary unit i is the number whose square is - 1.

i^2=- 1 In our case, the square root of a negative number is not real, but it can be written using the imaginary unit. This allows us to rewrite the given expression.

sqrt(- 49)

SqrtNegToSqrtI

sqrt(- a)= sqrt(a)* i

sqrt(49)* i

CalcRoot

Calculate root

We again want to calculate the value of the given expression. To do so, we can — similar as in Part A — use the fact that sqrt(a)* i = sqrt(- a).

sqrt(- 2)

SqrtNegToSqrtI

sqrt(- a)= sqrt(a)* i

sqrt(2) * i

CommutativePropMult

Commutative Property of Multiplication

isqrt(2)

We want to calculate the value of the given expression using the definition of the imaginary number i.

i^2=- 1 We can evaluate the expression using this definition.

(4i)^2

▼

Evaluate

PowProdII

(a b)^m=a^m b^m

4^2 i^2

CalcPow

Calculate power

16 i^2

IDefPow

i^2=- 1

16(-1)

MultPosNeg

a(- b)=- a * b

-16

Similarly, we want to evaluate the value of the given expression containing imaginary number i. To do so, first note that when the base of a power is a product, the expression can be rewritten by putting the exponent on both factors.

(3i)^3

PowProdII

(a b)^m=a^m b^m

3^3 i^3

CalcPow

Calculate power

27 i^3

To find the power of i, keep in mind that the Commutative and Associative Properties of Multiplication hold true for imaginary numbers. This allows us to rewrite the power as a product.

i^3 = i^2 * i Now — by the definition of imaginary unit — we know that i^2= - 1. i^2&= - 1 i^3&= i^2 * i= - 1 * i= - i Therefore, indeed i^3=- i. Finally, let's use this fact to simplify the given expression.

27i^3

Substitute

i^3= - i

27( - i)

MultPosNeg

a(- b)=- a * b

- 27i