The number i and Cartesian form Problem set A One question at a time

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

1. (a) \(i^{13}\)

   • \(i^{13} = (i^2)^{6} \times i = i\)

2. (b) \(-6i^{3}\)

   • \(i^{3} = -i\)
   • \(-6i^{3} = 6i\)

3. (c) \((4i)^2\)

   • \((4i)^2 = 16i^2 = -16\)

4. (d) \((2i)^3\)

   • \((2i)^3 = 8i^3 = 8 \times (-i)\)
   • \(= -8i\)

Answer each part.

1. (a) \(\sqrt{-81}\)

   • \(\sqrt{-81} = \sqrt{-1}\sqrt{81} = 9i\)

2. (b) \(\sqrt{-48}\)

   • \(\sqrt{-48} = \sqrt{48}\,i = 4 \sqrt{3}\,i\)

3. (c) State the real part and the imaginary part of \(5 - \sqrt{-16}\).

   • \(5 - \sqrt{-16} = 5 - 4i\)
   • \(\text{Re} = 5,\ \text{Im} = -4\)

Simplify each expression.

1. (a) \((5 - 2i) + (-3 + 9i)\)

   • \((5 + (-3)) + (-2 + (9))i = 2 + 7i\)

2. (b) \((-1 + 6i) - (4 - 3i)\)

   • \(-1 + 6i - (4 - 3i) = -1 + 6i - 4 + 3i\)
   • \(= -5 + 9i\)

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

1. (a) \((3 + 2i)(4 - 5i)\)

   • \((3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10i^2\)
   • \(= 22 - 7i\)

2. (b) \((-2 + i)(6 + 3i)\)

   • \((-2 + i)(6 + 3i) = -12 - 6i + 6i + 3i^2\)
   • \(= -15\)

Expand and simplify.

1. (a) \((2 + 7i)^2\)

   • \((2 + 7i)^2 = 4 + 28i + 49i^2\)
   • \(= -45 + 28i\)

2. (b) \((4 - i)^2\)

   • \((4 - i)^2 = 16 - 8i + i^2\)
   • \(= 15 - 8i\)

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

1. (a) \((3 - 2i)^2 - i(4 + 5i)\)

   • \((3 - 2i)^2 = 9 - 12i + 4i^2 = 5 - 12i\)
   • \(i(4 + 5i) = 4i + 5i^2 = -5 + 4i\)
   • \((5 - 12i) - (-5 + 4i) = 5 - 12i + 5 - 4i\)
   • \(= 10 - 16i\)

2. (b) \(2(1 + 3i)(2 - i) + 5i^3\)

   • \((1 + 3i)(2 - i) = 2 - i + 6i - 3i^2 = 5 + 5i\)
   • \(2(5 + 5i) = 10 + 10i, \quad 5i^3 = -5i\)
   • \(10 + 10i - 5i\)
   • \(= 10 + 5i\)

Use your calculator to check complex arithmetic.

1. (a) Set the calculator to complex mode. Evaluate \((6 - i)(2 + 5i)\) and write the result in the form \(x + iy\).

   • From the calculator: \(17 + 28i\)
   • Check by hand: \(12 + 30i - 2i - 5i^2 = 17 + 28i\)

2. (b) What does the calculator show for \(\sqrt{-1}\) in complex mode?

   • From the calculator: \(i\)
   • \(\sqrt{-1} = i\). In real mode it shows a non-real calculation message.

Karim writes \((3 + i)(3 - i) = 9 - i^2 = 8\). Dina writes the answer as \(10\).

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

   • \(9 - i^2 = 9 - (-1) = 9 + 1 = 10\)
   • Karim used \(i^2 = 1\). The correct value is \(i^2 = -1\).

2. (b) \((6 + 5i) - (2 + 9i)\) Simplify, and say why no \(i^2\) step is needed.

   • \(6 + 5i - (2 + 9i) = 6 + 5i - 2 - 9i\)
   • \(= 4 - 4i\)
   • Subtraction only combines real parts and imaginary parts, as in real algebra.

In an alternating current circuit, two components are in series. Their impedances are \(Z_1 = 3 + 4i\) ohms and \(Z_2 = 2 - 7i\) ohms. The total impedance is \(Z = Z_1 + Z_2\). The current is \(I = 2 - i\) amps, and the voltage is \(V = IZ\).

1. (a) Find the total impedance \(Z\).

   • \(Z = (3 + 2) + (4 - 7)i = 5 - 3i\)

2. (b) Find the voltage \(V\) in the form \(x + iy\).

   • \(V = (2 - i)(5 - 3i) = 10 - 6i - 5i + 3i^2\)
   • \(= 7 - 11i\)

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 \(i + i^2 + i^3 + \dots + i^{50}\).

   • The first 48 terms make 12 groups of four, each worth 0.
   • \(i^{49} + i^{50} = -1 + i\)

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 \((1 + i)^{8}\).

   • \((1 + i)^{8} = \left((1 + i)^2\right)^{4} = (2i)^{4}\)
   • \(= 16\)