Exercise 2.5 (Solutions)

  • Evaluate

(i) $i^7$ (ii) $i^{50}$ (iii) $i^{12}$ (iv) $\left(-i\right)^8$ (v) $\left(-i\right)^5$ (vi) $i^{27}$

Solution

(i) $$\begin{array}{cl} i^7 &= {i^6}\cdot i\\ &= (i^2)^3\cdot i\\ &= {-1}^3 \cdot i\\ &= -i \end{array}$$

(ii) $$\begin{array}{cl} i^{50} &= (i^2 )^{25}\\ &= {-1}^{25}\\ &= -1 \end{array}$$

(iii) $$\begin{array}{cl} i^{12} &= (i^2 )^6\\ &= {-1}^6\\ &= 1 \end{array}$$

(iv) $$\begin{array}{cl} (-i)^8 &= (-i^2 )^4\\ &= {-1}^4\\ &= 1 \end{array}$$

(v) $$\begin{array}{cl} (-i)^5 &= (-i)^4\cdot-i\\ &= ({-i}^2)^2\cdot-i\\ &= (-1)^2\cdot-i\\ &= -i \end{array}$$

(vi) $$\begin{array}{cl} i^{27} &= (i^2)^{13}\cdot i\\ &= {-1}^{13}\cdot i\\ &= (-1)\cdot i\\ &= -i \end{array}$$

  • Write the conjugate of the following numbers

(i) $2 + 3i$ (ii) $3 - 5i$ (iii) $-i$ (iv) $-3 + 4i$ (v) $-4 - i$ (vi) $i - 3$

Solution

(i) $$\begin{array}{cl} z = 2 + 3i\\ \bar{z} = 2 - 3i \end{array}$$

(ii) $$\begin{array}{cl} z = 3 - 5i\\ \bar{z} = 2 - 3i \end{array}$$

(iii) $$\begin{array}{cl} z = -i\\ \bar{z} = i \end{array}$$

(iv) $$\begin{array}{cl} z = -3 + 4i\\ \bar{z} = -3 - 4i \end{array}$$

(v) $$\begin{array}{cl} z = -4 - i\\ \bar{z} = -4 + i \end{array}$$

  • Write the real and imaginary part of the following numbers.
    • (i) $1 + i$
    • (ii) $-1 + 2i$
    • (iii) $-3i$ + 2$
    • (iv) $-2 - 2i$
    • (v) $-3i$
    • (vi) $2 + 0i$

Solution

(i) $$\begin{array}{cl} z = 1 + i\\ Re(z) = 1 ; Im(z) = 1 \end{array}$$

(ii) $$\begin{array}{cl} z = -1 + 2i\\ Re(z) = -1 ; Im(z) = 2 \end{array}$$

(iii) $$\begin{array}{cl} z = -3i + 2\\ Re(z) = 2 ; Im(z) = -3 \end{array}$$

(iv) $$\begin{array}{cl} z = -2 - 2i\\ Re(z) = -2 ; Im(z) = -2 \end{array}$$

(v) $$\begin{array}{cl} z = -3i\\ Re(z) = 0 ; Im(z) = -3 \end{array}$$

(vi) $$\begin{array}{cl} z = 2 + 0i\\ Re(z) = 2 ; Im(z) = 0 \end{array}$$

  • Find the value of x and y if

$x + iy + 1 = 4 - 3i$

Solution
$$\begin{array}{cl} x + iy + 1 = 4 - 3i\\ \hbox{ Saparate real and imaginary parts} \\ iy = -3i\\ \therefore y = -3\\ x + 1 = 4\\ x = 4 - 1\\ \therefore x = 3 \end{array}$$