Question

problem 7:
$(-6 + i)(-6 - i)$

problem 8:
$i^{23} =$

problem 9:
$i^7(1 + i^2)$

problem 10:
$i^6 + i^4 + i^2 + 1$

Explanation:

Problem 7

Step1: Use the formula \((a + b)(a - b)=a^2 - b^2\)

Here, \(a=-6\) and \(b = i\), so \((-6 + i)(-6 - i)=(-6)^2 - i^2\)

Step2: Calculate the values

We know that \(i^2=-1\) and \((-6)^2 = 36\). Substituting these values, we get \(36-(-1)=36 + 1=37\)

Step1: Recall the powers of \(i\)

We know that \(i^1 = i\), \(i^2=-1\), \(i^3=i^2\times i=-i\), \(i^4=(i^2)^2=(-1)^2 = 1\), and the powers of \(i\) repeat every 4. So we can divide the exponent by 4 and find the remainder.

Step2: Divide 23 by 4

\(23\div4 = 5\) with a remainder of 3. So \(i^{23}=i^{4\times5 + 3}=(i^4)^5\times i^3\)

Step3: Substitute the values

Since \(i^4 = 1\) and \(i^3=-i\), we have \((1)^5\times(-i)=-i\)

Step1: Simplify \(i^7\) and \(i^2\)

First, for \(i^7\), divide 7 by 4. \(7\div4 = 1\) with a remainder of 3. So \(i^7=i^{4\times1+3}=(i^4)^1\times i^3\). Since \(i^4 = 1\) and \(i^3=-i\), \(i^7=-i\). Also, \(i^2=-1\)

Step2: Substitute into the expression

Substitute \(i^7=-i\) and \(i^2=-1\) into \(i^7(1 + i^2)\), we get \(-i(1+(-1))\)

Step3: Simplify the expression

Simplify inside the parentheses first: \(1+(-1)=0\). Then \(-i\times0 = 0\)

Answer:

\(37\)

Problem 8