Exercise 1.1

1. Solve the following equations.

(a) Solution

$$ x^2 - 6x + 10 = 0 $$ $$ x^2 - 6x + 9 = -1 $$ $$ (x - 3)^2 = i^2 $$ $$ x - 3 = \pm i $$ $$ x = 3 \pm i $$

So there are two solutions: \( x = 3 + i \) and \( x = 3 - i \).

(b) Solution

$$ -2x^2 + 4x - 3 = 0 $$ $$ x^2 - 2x + \frac{3}{2} = 0 $$ $$ x^2 - 2x + 1 = -\frac{1}{2} $$ $$ (x - 1)^2 = \frac{1}{2}i^2 $$ $$ x - 1 = \pm \sqrt{\frac{1}{2}}i $$ $$ x = 1 \pm \frac{\sqrt{2}}{2}i $$

So there are two solutions: \( x = 1 + \frac{\sqrt{2}}{2}i \) and \( x = 1 - \frac{\sqrt{2}}{2}i \).

(c) Solution

$$ 5x^2 - 2x + 1 = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{5} = 0 $$ $$ x^2 - \frac{2}{5}x + \frac{1}{25} = -\frac{4}{25} $$ $$ (x - \frac{1}{5})^2 = \frac{4}{25}i^2 $$ $$ x - \frac{1}{5} = \pm \frac{2}{5}i $$ $$ x = \frac{1}{5} \pm \frac{2}{5}i $$

So there are two solutions: \( x = \frac{1}{5} + \frac{2}{5}i \) and \( x = \frac{1}{5} - \frac{2}{5}i \).

(d) Solution

$$ 3x^2 + 7x + 5 = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{5}{3} = 0 $$ $$ x^2 + \frac{7}{3}x + \frac{49}{36} = -\frac{11}{36} $$ $$ (x + \frac{7}{6})^2 = \frac{11}{36}i^2 $$ $$ x + \frac{7}{6} = \pm \frac{\sqrt{11}}{6}i $$ $$ x = -\frac{7}{6} \pm \frac{\sqrt{11}}{6}i $$

So there are two solutions: \( x = -\frac{7}{6} + \frac{\sqrt{11}}{6}i \) and \( x = -\frac{7}{6} - \frac{\sqrt{11}}{6}i \).

2. Solve the following equations and check your answers.

(a) Solution

$$x^2 - 2x + 4 = 0$$ $$x^2 - 2x + 1 = -3$$ $$(x - 1)^2 = 3i^2$$ $$x - 1 = \pm \sqrt{3}i$$ $$x = 1 \pm \sqrt{3}i$$

So the two solutions are \( x = 1 + \sqrt{3}i \) and \( x = 1 - \sqrt{3}i \).

For \(x = 1 + \sqrt{3}i\),

\((1 + \sqrt{3}i)^2 - 2(1 + \sqrt{3}i) + 4\)

\(= (1 + 2\sqrt{3}i - 3) - 2 - 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

For \(x = 1 - \sqrt{3}i\),

\((1 - \sqrt{3}i)^2 - 2(1 - \sqrt{3}i) + 4\)

\(= (1 - 2\sqrt{3}i - 3) - 2 + 2\sqrt{3}i + 4\)

\(= -2 - 2 + 4 = 0\)

(b) Solution

$$x^2 - 4x + 5 = 0$$ $$x^2 - 4x + 4 = -1$$ $$(x - 2)^2 = i^2$$ $$x - 2 = \pm i$$ $$x = 2 \pm i$$

So the two solutions are \( x = 2 + i \) and \( x = 2 - i \).

For \(x = 2 + i\),

\((2+i)^2 - 4(2+i) + 5\)

\(= (4 + 4i - 1) - 8 - 4i + 5\)

\(= 3 - 8 + 5 = 0\)

For \(x = 2 - i\),

\((2-i)^2 - 4(2-i) + 5\)

\(= (4 - 4i - 1) - 8 + 4i + 5\)

\(= 3 - 8 + 5 = 0\)

3. Find the value of \(i^n\) for every positive integer \(n\).

Solution (Method 1)

The powers of \(i\) follow a repeating cycle. We can find this pattern by calculating the first few powers step-by-step, using the definition \(i^2 = -1\).

• Step 1: Calculate \(i^2\).
  \(i^2 = -1\) (by definition)
• Step 2: Calculate \(i^3\).
  \(i^3 = i^2 \cdot i = (-1) \cdot i = -i\)
• Step 3: Calculate \(i^4\).
  \(i^4 = i^2 \cdot i^2 = (-1)(-1) = 1\)
• Step 4: Calculate \(i^5\).
  \(i^5 = i^4 \cdot i = (1) \cdot i = i\)

The pattern \(i, -1, -i, 1\) begins to repeat. This means the value of \(i^n\) depends on which category the exponent \(n\) falls into. We can express this using an integer \(k\).

• If \(n\) is of the form \(4k+1\) (e.g., 1, 5, 9,...), then \(i^n = i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+2\) (e.g., 2, 6, 10,...), then \(i^n = -1\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k+3\) (e.g., 3, 7, 11,...), then \(i^n = -i\). (for \(k \ge 0\))
• If \(n\) is of the form \(4k\) (e.g., 4, 8, 12,...), then \(i^n = 1\). (for \(k \ge 1\))

This can be written as a piecewise-defined function where \(k\) is an integer satisfying the conditions above:

$$ i^n = \begin{cases} i & \text{if } n = 4k+1 \\ -1 & \text{if } n = 4k+2 \\ -i & \text{if } n = 4k+3 \\ 1 & \text{if } n = 4k \end{cases} $$

Solution (Method 2: )

An alternative method is to consider whether the exponent \(n\) is even or odd.

If \(n\) is an even integer, we can write \(n=2k\). Then \(i^n = i^{2k} = (i^2)^k = (-1)^k\). Since \(k=\frac{n}{2}\), this gives \(i^n = (-1)^{\frac{n}{2}}\).

If \(n\) is an odd integer, we can write \(n=2k+1\). Then \(i^n = i^{2k+1} = i^{2k} \cdot i = (-1)^k \cdot i\). Since \(k=\frac{n-1}{2}\), this gives \(i^n = (-1)^{\frac{n-1}{2}}i\).

This can be summarized in the following piecewise-defined function:

$$ i^n = \begin{cases} (-1)^{\frac{n}{2}} & \text{if } n \text{ is even} \\ (-1)^{\frac{n-1}{2}}i & \text{if } n \text{ is odd}\end{cases}$$

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