For the imaginary unit i, where i^2 = -1, the complex number z is defined by: z = \frac{1 … — ACT

For the imaginary unit $i$ , where $i^2 = -1$ , the complex number $z$ is defined by:

z = \frac{1 - 3i^{45}}{1 + i^{95}}

Which of the following is equivalent to $z^2$ ?

3 - 4i

The correct answer is found by first simplifying the powers of

i

in the expression for

z

i^{45} = i

and

i^{95} = -i

, which yields

z = \frac{1 - 3i}{1 - i}

. Rationalizing the fraction by multiplying both the numerator and the denominator by the conjugate of the denominator (

1 + i

) simplifies the expression to

z = 2 - i

. Finally, squaring this result using the binomial expansion formula gives

(2 - i)^2 = 4 - 4i + i^2 = 3 - 4i

Simplify the high integer powers of the imaginary unit

i

by using the fact that powers of

i

repeat in a cycle of four:

i^1 = i

i^2 = -1

i^3 = -i

, and

i^4 = 1

i^{45} = i

and

i^{95} = -i

Since

45 = 4(11) + 1

, the remainder is

1

, so

i^{45} = i^1 = i

. Since

95 = 4(23) + 3

, the remainder is

3

, so

i^{95} = i^3 = -i

Substitute these simplified values back into the expression for

z

z = \frac{1 - 3i}{1 - i}

This sets up the fraction with simplified imaginary terms in both the numerator and the denominator.

Rationalize the denominator by multiplying the numerator and denominator of the fraction by the complex conjugate of the denominator, which is

1 + i

z = 2 - i

Multiplying by the conjugate eliminates the imaginary unit from the denominator:

\frac{(1 - 3i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + i - 3i - 3i^2}{1 - i^2} = \frac{4 - 2i}{2} = 2 - i

Calculate the value of

z^2

by squaring the simplified complex number

2 - i

z^2 = 3 - 4i

Using the binomial squaring formula

(a - b)^2 = a^2 - 2ab + b^2

, we expand

(2 - i)^2

to get

2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i

Simplifying complex numbers by evaluating powers of

i

, rationalizing fractions with complex conjugates, and expanding complex binomials.

Tahmini Süre:1m 30s

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