Generate Quiz 6A40D307DE626

In response, b (x=1, y=0)

[1-1+ii] is the solution.100= 1-i1 +i ×1-i1-i100= 1-i(1-i)(1+i)(1-i)100=(1-i)12 -21i+ i212-i2100=1-2i+(-1)1-(-1)100 = 212-i2100

=1-2i-11+1100= -2i 2100= -i100=-i250=-(-1)50= 1

In x+iy form 1+0i

This is only feasible with zero since additive identity is the property that when a number is added to another number, the original number is returned.

The ability to eliminate common factors on opposing sides of an equation under specific circumstances is known as the cancellation property. We can observe that “a” shows up on both sides of the equation in this instance.

To understand how cancellation works, let’s simplify the equation:

Ra + c = b + cab

Taking “c” away from each side:

Ra = b + cab – c

Taking “a” out of the right side:

Ra = b + a(c – b)

Only when a ≠ 0 can we divide both sides by “a” because we are working with real numbers (R). The primary requirement for cancellation is this. It is not defined to divide by zero.

Dividing both sides by a, assuming a ≠ 0, yields:

R = b/a + (c – b)

Consequently, cancellation permits R to be isolated provided that a ≠ 0.

Non-negative integer: Any whole number that is zero or higher than zero is considered a non-negative integer. It fits this definition because 0 is zero.

Any number that can be shown on a number line, including fractions, decimals, and positive and negative numbers, is considered a real number. Zero is a real number that lies on the number line.

Whole numbers are integers that are not negative (0, 1, 2, 3,…). The set of whole numbers includes zero.

Consequently, 0 is a whole number, a real number, and a non-negative integer. Therefore, every option is relevant.

The number that, when multiplied by any other number, does not change is the multiplicative identity of any number system. The multiplicative identity for complex numbers, or numbers of the form a + bi, where a and b are real numbers and i is the imaginary unit, is (1, 0).

decimal that is neither recurring nor terminating.

Knowing the characteristics of the imaginary unit “i” and integral multiples of 4 is essential to answering this question.

Integral multiple of 4: Any number that results from multiplying 4 by an integer is an integral multiple of 4. Put differently, 4n is an integral multiple of 4 (e.g., 4, 8, 12, -4, -8, etc.) if n is an integer.

Imaginary unit “i”: The square root of -1 is the definition of the imaginary unit “i,” so i2 = -1.

The right choice is option D.

This is the reason:

Assuming n = 4k (where k is another integer) and n is an integral multiple of 4, then:

in = i(4k)

A property of exponents states that (am)n = a(m.n). Consequently:

(i4)k = i(4k)

i2 = -1 is known to us. Thus:

i4 = (i2)2 = (-1)2 = 1.

Consequently:

(i4)k = 1k = 1 = i(4k)

This implies that i raised to the power of n will be 1 if n is a multiple of 4. We can also express it as ±i, though, since i is the square root of -1, and squaring either +i or -i yields 1.