Question

simplify the expression to a + bi form: $3i^{5}+9i^{21}-12i^{46}+8i^{105}$

Explanation:

Step1: Recall the powers of \(i\)

We know that \(i^1 = i\), \(i^2=-1\), \(i^3 = i^2\cdot i=-i\), \(i^4=(i^2)^2 = (-1)^2 = 1\), and the powers of \(i\) repeat every 4. So we can find the remainder when the exponent is divided by 4 to simplify \(i^n\).

Step2: Simplify \(i^5\)

For \(i^5\), divide 5 by 4: \(5\div4 = 1\) with a remainder of 1. So \(i^5=i^{4 + 1}=i^4\cdot i^1\). Since \(i^4 = 1\), then \(i^5 = 1\cdot i=i\). So \(3i^5=3i\).

Step3: Simplify \(i^{21}\)

For \(i^{21}\), divide 21 by 4: \(21\div4 = 5\) with a remainder of 1. So \(i^{21}=i^{4\times5+1}=(i^4)^5\cdot i^1\). Since \(i^4 = 1\), \((i^4)^5 = 1^5 = 1\), so \(i^{21}=1\cdot i = i\). Thus, \(9i^{21}=9i\).

Step4: Simplify \(i^{46}\)

For \(i^{46}\), divide 46 by 4: \(46\div4 = 11\) with a remainder of 2. So \(i^{46}=i^{4\times11 + 2}=(i^4)^{11}\cdot i^2\). Since \(i^4 = 1\), \((i^4)^{11}=1^{11}=1\), and \(i^2=-1\), so \(i^{46}=1\times(-1)=-1\). Thus, \(- 12i^{46}=-12\times(-1)=12\).

Step5: Simplify \(i^{105}\)

For \(i^{105}\), divide 105 by 4: \(105\div4 = 26\) with a remainder of 1. So \(i^{105}=i^{4\times26+1}=(i^4)^{26}\cdot i^1\). Since \(i^4 = 1\), \((i^4)^{26}=1^{26}=1\), so \(i^{105}=1\cdot i = i\). Thus, \(8i^{105}=8i\).

Step6: Combine like terms

Now we substitute the simplified terms back into the original expression:
\(3i^5 + 9i^{21}-12i^{46}+8i^{105}=3i + 9i+12 + 8i\)
Combine the \(i\) terms: \((3i + 9i+8i)+12=(3 + 9 + 8)i+12 = 20i+12\)
We can write it in the form \(a + bi\) as \(12+20i\).

Answer:

\(12 + 20i\)