Algebra 2 Complex Conjugates and Complex Solutions Worksheet Solutions (PDF)

Change the sign of the imaginary part: \(4 – 3i\) becomes \(4 + 3i\).

\(4 + 3i\)

\(6i = 0 + 6i\) has conjugate \(0 – 6i = -6i\). The real number \(-9 = -9 + 0i\) is its own conjugate.

\(-6i\) and \(-9\)

\(9 – 15i + 15i – 25i^{2} = 9 + 25 = 34\). This equals \(a^{2} + b^{2} = 9 + 25\).

\(34\)

\(|3 + 4i| = \sqrt{3^{2} + 4^{2}} = \sqrt{25}\).

\(5\)

\(\sqrt{2^{2} + (-4)^{2}} = \sqrt{20} = \sqrt{4 \cdot 5}\).

\(2\sqrt{5}\)

Multiply the numerator and denominator by the conjugate of the denominator, \(1 – i\):

\(\frac{(6)(1 – i)}{(1 + i)(1 – i)}\)

Numerator: \(6 – 6i\). Denominator: \(1 + 1 = 2\).

Divide each part by \(2\): \(3 – 3i\).

\(3 – 3i\)

Multiply the numerator and denominator by the conjugate of the denominator, \(2 + i\):

\(\frac{(10 + 5i)(2 + i)}{(2 – i)(2 + i)}\)

Numerator: \(15 + 20i\). Denominator: \(4 + 1 = 5\).

Divide each part by \(5\): \(3 + 4i\).

\(3 + 4i\)

Multiply the numerator and denominator by the conjugate of the denominator, \(3 – i\):

\(\frac{(5i)(3 – i)}{(3 + i)(3 – i)}\)

Numerator: \(5 + 15i\). Denominator: \(9 + 1 = 10\).

Divide each part by \(10\): \(\frac{1}{2} + \frac{3}{2}i\).

\(\frac{1}{2} + \frac{3}{2}i\)

Multiply the numerator and denominator by the conjugate of the denominator, \(1 + 4i\):

\(\frac{(3 + 2i)(1 + 4i)}{(1 – 4i)(1 + 4i)}\)

Numerator: \(-5 + 14i\). Denominator: \(1 + 16 = 17\).

Divide each part by \(17\): \(-\frac{5}{17} + \frac{14}{17}i\).

\(-\frac{5}{17} + \frac{14}{17}i\)

The conjugate of \(i\) is \(-i\). Multiply top and bottom by \(-i\): \(\frac{(4 + 3i)(-i)}{i(-i)} = \frac{-4i – 3i^{2}}{-i^{2}} = \frac{-4i + 3}{1}\).

\(3 – 4i\)

Identify \(a = 1, b = 4, c = 13\).

Discriminant: \(b^{2} – 4ac = (4)^{2} – 4(1)(13) = -36\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{-4 \pm \sqrt{-36}}{2}\) with \(\sqrt{-36} = 6i\).

\(x = \frac{-4 \pm 6i}{2} = -2 \pm 3i\).

\(-2 + 3i\) and \(-2 – 3i\)

Identify \(a = 1, b = -6, c = 10\).

Discriminant: \(b^{2} – 4ac = (-6)^{2} – 4(1)(10) = -4\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{6 \pm \sqrt{-4}}{2}\) with \(\sqrt{-4} = 2i\).

\(x = \frac{6 \pm 2i}{2} = 3 \pm i\).

\(3 + i\) and \(3 – i\)

Identify \(a = 2, b = 2, c = 5\).

Discriminant: \(b^{2} – 4ac = (2)^{2} – 4(2)(5) = -36\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{-2 \pm \sqrt{-36}}{4}\) with \(\sqrt{-36} = 6i\).

\(x = \frac{-2 \pm 6i}{4} = -\frac{1}{2} \pm \frac{3}{2}i\).

\(-\frac{1}{2} + \frac{3}{2}i\) and \(-\frac{1}{2} – \frac{3}{2}i\)

\(5x^{2} = -20\), so \(x^{2} = -4\).

\(x = \pm\sqrt{-4} = \pm 2i\).

\(x = 2i\) or \(x = -2i\)

The graph never touches the x-axis, so there are 0 real solutions.

Identify \(a = 1, b = -4, c = 6\).

Discriminant: \(b^{2} – 4ac = (-4)^{2} – 4(1)(6) = -8\).

Since the discriminant is negative, the solutions are complex: \(x = \frac{4 \pm \sqrt{-8}}{2}\) with \(\sqrt{-8} = 2\sqrt{2}i\).

\(\sqrt{-8} = 2i\sqrt{2}\), so \(x = \frac{4 \pm 2i\sqrt{2}}{2} = 2 \pm i\sqrt{2}\).

No real solutions; \(x = 2 \pm i\sqrt{2}\)

Sum of roots: \((2 + 3i) + (2 – 3i) = 4\). Product: \((2 + 3i)(2 – 3i) = 4 + 9 = 13\).

A monic quadratic is \(x^{2} – (\text{sum})x + \text{product}\).

\(x^{2} – 4x + 13 = 0\)

\(-16t^{2} + 32t = 20\) becomes \(16t^{2} – 32t + 20 = 0\), which simplifies to \(4t^{2} – 8t + 5 = 0\).

Discriminant: \((-8)^{2} – 4(4)(5) = 64 – 80 = -16\), which is negative.

The solutions are complex: \(t = \frac{8 \pm 4i}{8} = 1 \pm \frac{1}{2}i\). No real time works, so the ball never reaches 20 feet. Its maximum height is \(h(1) = 16\) feet.

Maya is wrong; the maximum is 16 ft

Multiply the numerator and denominator by the conjugate of the denominator, \(3 – i\):

\(\frac{(10 + 20i)(3 – i)}{(3 + i)(3 – i)}\)

Numerator: \(50 + 50i\). Denominator: \(9 + 1 = 10\).

Divide each part by \(10\): \(5 + 5i\).

\(5 + 5i\) ohms

Multiply by \(\frac{1 + 2i}{1 + 2i}\): numerator \((1 + 2i)^{2} = 1 + 4i – 4 = -3 + 4i\), denominator \(1 + 4 = 5\).

\(-\frac{3}{5} + \frac{4}{5}i\)

\(z_{0}^{2} = (1 + i)^{2} = 1 + 2i – 1 = 2i\).

\(z_{1} = 2i + (-1 + 2i) = -1 + 4i\).

\(|z_{1}| = \sqrt{1 + 16} = \sqrt{17}\).

\(z_{1} = -1 + 4i\), \(|z_{1}| = \sqrt{17}\)

Complex solutions of a real quadratic come in conjugate pairs, so the other solution is \(3 + 2i\).

Sum \(= 6\), product \(= 9 + 4 = 13\), so \(b = -6\) and \(c = 13\).

\(3 + 2i\); \(x^{2} – 6x + 13 = 0\)