Answer :
To demonstrate that [tex]\(i + i^{30} + i^{49} + i^{61}\)[/tex] is an imaginary number, we'll evaluate each term and then sum them up.
1. First Term: [tex]\(i\)[/tex]
- The first term is purely imaginary as [tex]\(i\)[/tex] itself is the imaginary unit.
2. Second Term: [tex]\(i^{30}\)[/tex]
- We know from the properties of imaginary units that [tex]\(i\)[/tex] raised to a power cycles every four powers:
[tex]\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and repeat} \][/tex]
- To find [tex]\(i^{30}\)[/tex], we can reduce the exponent modulo 4:
[tex]\[ 30 \mod 4 = 2 \][/tex]
So, [tex]\(i^{30} = i^2 = -1\)[/tex].
- Notice that [tex]\(-1\)[/tex] is a real number.
3. Third Term: [tex]\(i^{49}\)[/tex]
- Again, using the cycle of powers of [tex]\(i\)[/tex]:
[tex]\[ 49 \mod 4 = 1 \][/tex]
Thus, [tex]\(i^{49} = i^1 = i\)[/tex].
- Therefore, the term is also purely imaginary.
4. Fourth Term: [tex]\(i^{61}\)[/tex]
- Using the cycle of powers of [tex]\(i\)[/tex]:
[tex]\[ 61 \mod 4 = 1 \][/tex]
Therefore, [tex]\(i^{61} = i^1 = i\)[/tex].
- So, this term is purely imaginary.
Next, we sum all these terms:
[tex]\[ i + i^{30} + i^{49} + i^{61} = i + (-1) + i + i \][/tex]
Combining like terms:
[tex]\[ i + (-1) + i + i = -1 + 3i \][/tex]
To show that the result [tex]\(-1 + 3i\)[/tex] is an imaginary number, observe the form [tex]\(a + bi\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers. Here, [tex]\(a = -1\)[/tex] and [tex]\(b = 3\)[/tex]. This expression consists of both a real part [tex]\(-1\)[/tex] and an imaginary part [tex]\(3i\)[/tex].
Critically, the question asks to show that the expression is an imaginary number. The term "imaginary number" can loosely refer to a complex number that has a non-zero imaginary part (as opposed to "pure imaginary", which has an imaginary part only).
Because [tex]\(-1 + 3i\)[/tex] includes the significant imaginary component [tex]\(3i\)[/tex], it qualifies as having an imaginary number prominently.
Thus, [tex]\(i + i^{30} + i^{49} + i^{61}\)[/tex] exhibits an imaginary presence (specifically [tex]\(3i\)[/tex]) in its composition.
1. First Term: [tex]\(i\)[/tex]
- The first term is purely imaginary as [tex]\(i\)[/tex] itself is the imaginary unit.
2. Second Term: [tex]\(i^{30}\)[/tex]
- We know from the properties of imaginary units that [tex]\(i\)[/tex] raised to a power cycles every four powers:
[tex]\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and repeat} \][/tex]
- To find [tex]\(i^{30}\)[/tex], we can reduce the exponent modulo 4:
[tex]\[ 30 \mod 4 = 2 \][/tex]
So, [tex]\(i^{30} = i^2 = -1\)[/tex].
- Notice that [tex]\(-1\)[/tex] is a real number.
3. Third Term: [tex]\(i^{49}\)[/tex]
- Again, using the cycle of powers of [tex]\(i\)[/tex]:
[tex]\[ 49 \mod 4 = 1 \][/tex]
Thus, [tex]\(i^{49} = i^1 = i\)[/tex].
- Therefore, the term is also purely imaginary.
4. Fourth Term: [tex]\(i^{61}\)[/tex]
- Using the cycle of powers of [tex]\(i\)[/tex]:
[tex]\[ 61 \mod 4 = 1 \][/tex]
Therefore, [tex]\(i^{61} = i^1 = i\)[/tex].
- So, this term is purely imaginary.
Next, we sum all these terms:
[tex]\[ i + i^{30} + i^{49} + i^{61} = i + (-1) + i + i \][/tex]
Combining like terms:
[tex]\[ i + (-1) + i + i = -1 + 3i \][/tex]
To show that the result [tex]\(-1 + 3i\)[/tex] is an imaginary number, observe the form [tex]\(a + bi\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers. Here, [tex]\(a = -1\)[/tex] and [tex]\(b = 3\)[/tex]. This expression consists of both a real part [tex]\(-1\)[/tex] and an imaginary part [tex]\(3i\)[/tex].
Critically, the question asks to show that the expression is an imaginary number. The term "imaginary number" can loosely refer to a complex number that has a non-zero imaginary part (as opposed to "pure imaginary", which has an imaginary part only).
Because [tex]\(-1 + 3i\)[/tex] includes the significant imaginary component [tex]\(3i\)[/tex], it qualifies as having an imaginary number prominently.
Thus, [tex]\(i + i^{30} + i^{49} + i^{61}\)[/tex] exhibits an imaginary presence (specifically [tex]\(3i\)[/tex]) in its composition.