Question

level 4: open ended questions

1. find two complex numbers that have a sum of 10i, a difference of -4 and a product of -29
2. complete the table

	process
$i^1$	$i$
$i^2$	-1
$i^3$	$i^2cdot i$	-$i$
$i^4$	$(i^2)^2$	1
$i^5$	$i^4cdot i$	$i$
$i^6$	$i^4cdot i^2$	-1
$i^7$	$i^4cdot i^3$	-$i$
$i^8$	$(i^4)^2$	1

Explanation:

Step1: Let the two complex numbers be \(a + bi\) and \(c+di\)

Let \(z_1=a + bi\) and \(z_2=c + di\). We know that \(z_1 + z_2=(a + c)+(b + d)i = 10i\), \(z_1 - z_2=(a - c)+(b - d)i=-4\) and \(z_1z_2=(a + bi)(c + di)=(ac - bd)+(ad+bc)i=-29\).
From \(z_1 + z_2=(a + c)+(b + d)i = 10i\), we have \(a + c = 0\) and \(b + d=10\). From \(z_1 - z_2=(a - c)+(b - d)i=-4\), we have \(a - c=-4\) and \(b - d = 0\).

Step2: Solve the system of equations for \(a\) and \(c\)

Solve \(

$$\begin{cases}a + c = 0\\a - c=-4\end{cases}$$

\). Add the two equations: \((a + c)+(a - c)=0+( - 4)\), \(2a=-4\), so \(a=-2\) and \(c = 2\).

Step3: Solve the system of equations for \(b\) and \(d\)

Solve \(

$$\begin{cases}b + d=10\\b - d = 0\end{cases}$$

\). Add the two equations: \((b + d)+(b - d)=10 + 0\), \(2b=10\), so \(b = 5\) and \(d = 5\).
So the two complex numbers are \(-2+5i\) and \(2 + 5i\).
For the table - the powers of \(i\) follow a cycle.

Step4: Analyze \(i^8\)

Since \(i^4 = 1\), and \(8=4\times2\), so \(i^8=(i^4)^2=1^2 = 1\)

Answer:

The two complex numbers are \(-2 + 5i\) and \(2+5i\). The value of \(i^8\) is \(1\).