The number i and Cartesian form Problem set B Worked solutions

Simplify each expression, giving the answer in the form \(x + iy\).

1. (a) \(i^{18}\)

   • \(i^{18} = (i^2)^{9} = (-1)^{9} = -1\)

2. (b) \(5i^{7}\)

   • \(i^{7} = -i\)
   • \(5i^{7} = -5i\)

3. (c) \((3i)^2\)

   • \((3i)^2 = 9i^2 = -9\)

4. (d) \((5i)^3\)

   • \((5i)^3 = 125i^3 = 125 \times (-i)\)
   • \(= -125i\)

Answer each part.

1. (a) \(\sqrt{-64}\)

   • \(\sqrt{-64} = \sqrt{-1}\sqrt{64} = 8i\)

2. (b) \(\sqrt{-75}\)

   • \(\sqrt{-75} = \sqrt{75}\,i = 5 \sqrt{3}\,i\)

3. (c) State the real part and the imaginary part of \(-2 - \sqrt{-49}\).

   • \(-2 - \sqrt{-49} = -2 - 7i\)
   • \(\text{Re} = -2,\ \text{Im} = -7\)

Simplify each expression.

1. (a) \((-4 + 7i) + (8 - 3i)\)

   • \((-4 + (8)) + (7 + (-3))i = 4 + 4i\)

2. (b) \((6 - 2i) - (-5 + 4i)\)

   • \(6 - 2i - (-5 + 4i) = 6 - 2i + 5 - 4i\)
   • \(= 11 - 6i\)

Expand and simplify, giving each answer in the form \(x + iy\).

1. (a) \((2 - 3i)(5 + i)\)

   • \((2 - 3i)(5 + i) = 10 + 2i - 15i - 3i^2\)
   • \(= 13 - 13i\)

2. (b) \((4 + i)(-3 - 2i)\)

   • \((4 + i)(-3 - 2i) = -12 - 8i - 3i - 2i^2\)
   • \(= -10 - 11i\)

Expand and simplify.

1. (a) \((5 - 3i)^2\)

   • \((5 - 3i)^2 = 25 - 30i + 9i^2\)
   • \(= 16 - 30i\)

2. (b) \((1 + 6i)^2\)

   • \((1 + 6i)^2 = 1 + 12i + 36i^2\)
   • \(= -35 + 12i\)

Simplify each expression, giving the answer in the form \(x + iy\).

1. (a) \((1 + 4i)^2 + 2i(3 - i)\)

   • \((1 + 4i)^2 = 1 + 8i + 16i^2 = -15 + 8i\)
   • \(2i(3 - i) = 6i - 2i^2 = 2 + 6i\)
   • \((-15 + 8i) + (2 + 6i)\)
   • \(= -13 + 14i\)

2. (b) \(3(2 - i)(1 + 2i) - 4i^3\)

   • \((2 - i)(1 + 2i) = 2 + 4i - i - 2i^2 = 4 + 3i\)
   • \(3(4 + 3i) = 12 + 9i, \quad -4i^3 = 4i\)
   • \(12 + 9i + 4i\)
   • \(= 12 + 13i\)

Use your calculator to check complex arithmetic.

1. (a) Set the calculator to complex mode. Evaluate \((3 + 4i)(5 - 2i)\) and write the result in the form \(x + iy\).

   • From the calculator: \(23 + 14i\)
   • Check by hand: \(15 - 6i + 20i - 8i^2 = 23 + 14i\)

2. (b) What does the calculator show for \(\sqrt{-25}\) in complex mode?

   • From the calculator: \(5i\)
   • \(\sqrt{-25} = 5i\). In real mode it shows a non-real calculation message.

Ana writes \((5 + 2i)(5 - 2i) = 25 - 4i^2 = 21\). Ben writes the answer as \(29\).

1. (a) Which of them is right? Explain the mistake.

   • \(25 - 4i^2 = 25 - 4(-1) = 25 + 4 = 29\)
   • Ana used \(i^2 = 1\). The correct value is \(i^2 = -1\).

2. (b) \((-1 + 4i) - (2 + 3i)\) Simplify, and say why no \(i^2\) step is needed.

   • \(-1 + 4i - (2 + 3i) = -1 + 4i - 2 - 3i\)
   • \(= -3 + i\)
   • Subtraction only combines real parts and imaginary parts, as in real algebra.

Two sensors report a signal as complex numbers, in millivolts: \(p = 4 - 3i\) and \(q = -1 + 8i\). The combined signal is \(s = p + q\). A filter multiplies the combined signal by the gain \(g = 3 + i\), so the output is \(gs\).

1. (a) Find the combined signal \(s\).

   • \(s = (4 - 1) + (-3 + 8)i = 3 + 5i\)

2. (b) Find the output \(gs\) in the form \(x + iy\).

   • \(gs = (3 + i)(3 + 5i) = 9 + 15i + 3i + 5i^2\)
   • \(= 4 + 18i\)

The powers of i repeat in a cycle of four.

1. (a) Find \(i + i^2 + i^3 + i^4\).

   • \(i + (-1) + (-i) + 1 = 0\)

2. (b) Hence find \(i + i^2 + i^3 + \dots + i^{47}\).

   • The first 44 terms make 11 groups of four, each worth 0.
   • \(i^{45} + i^{46} + i^{47} = -1\)

This question uses a square to find a high power.

1. (a) Show that \((1 - i)^2 = -2i\).

   • \((1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i\)

2. (b) Hence find \((1 - i)^{6}\).

   • \((1 - i)^{6} = \left((1 - i)^2\right)^{3} = (-2i)^{3}\)
   • \(= 8i\)