# Exercise 2.6 (Solutions)

### Question 1

Identify the following statements as true or false. (i) $\sqrt{-3}\cdot\sqrt{-3} = 3$
(ii) $i^{73}=-i$
(iii) $i^{10} = -1$
(iv) Complex conjugate of $(-6i + i^2) is (-1 + 6i)$
(v) Difference of complex numbers $z = a + ib$ and its conjugate is a real number.
(vi) If $(a-1)-(b+3)i = 5+8i$, then a = 6 & b = -11
(vii) Product of complex number and its conjugate is always a non-negative real number.

Solution

(i) False

(ii) False

(iii) True

(iv) True

(v) False

(vi) True

(vii) True

### Question 2

Express each complex number in the standrad form $a+ib$, where a and b are real numbers. (i) $(2+3i)+(7-2i)$

(ii) $2(5+4i)+3(7-4i)$

(iii) $-(-3+5i)-3(4+9i)$

(iv) $2i^2+6i^3+3i^{16}-6i^{19}+4i^{25}$

Solution

(i) $$\begin{array}{cl} (2+3i)+(7-2i) &= 2+3i+7-2i\\ &= 2+7+3i-2i\\ &= 9+i \end{array}$$

(ii) $$\begin{array}{cl} 2(5+4i)-3(7+4i) &= 10+8i-21-12i\\ &= 10-21+8i-12i\\ &= -11-4i \end{array}$$

(iii) $$\begin{array}{cl} -(-3+5i)-(4+9i) &= 3-5i-4-9i\\ &= 3-4-5i-9i\\ &= -1-14i \end{array}$$

(iv) $$\begin{array}{cl} 2i^2+6i^3+3i^{16}-6i^{19}+4i^{25} &= 2(-1)+6(-1)(i)+3(i^2)^{\times8}-6(i^2)^{\times9}(i)+4(i^2)^{\times12}(i)\\ &= -2 -6i+3+6i+4i\\ &= 1+4i \end{array}$$

### Question 3

• Simplify and write your answer in the form of $a+ib$

(i) $(-7+3i)(-3+2i)$
(ii) $(2-{\sqrt{-4}})(3-{\sqrt{-4}})$
(iii) $({\sqrt{5}}-3i)^2$
(iv) $(2-3i){\overline{(3-2i)}}$

Solution

(i) $$\begin{array}{cl} (-7+3i)(-3+2i) &= -7(-3+2i)+3i(-3+2i) &= 21-14i-9i+6i^2\\ &= 21-23i+6(-1)\\ &= 21-6-23i\\ &= 15-23i \end{array}$$

(ii) $$\begin{array}{cl} (2-\sqrt{-4})(3-\sqrt{-4})\\ &= (2-2i)(3-2i)\\ &= 2(3-2i)-2i(3-2i)\\ &= 6-4i-6i+4i^2\\ &= 6-10i+4(-1)\\ &= 6-4-10i\\ &= 2-10i \end{array}$$

(iii) $$\begin{array}{cl} (\sqrt{5}-3i)^2 &= (\sqrt{5}-3i)(\sqrt{5}-3i)\\ &= \sqrt{5}({\sqrt{5}}-3i)-3i(\sqrt{5}-3i)\\ &= 5-3\sqrt{5}i-3\sqrt{5}i+9i^2\\ &= 5-6\sqrt{5}i+9(-1)\\ &= 5-9-6\sqrt{5}i\\ &= -4-6\sqrt{5}i \end{array}$$

(iv) $$\begin{array}{cl} (2-3i){\overline{(3-2i)}}\\ &= (2-3i)(3+2i)\\ &= 2(3+2i)-3i(3+2i)\\ &= 6+4i-9i-6i^2\\ &= 6-5i-6(-1)\\ &= 6+6-5i\\ &= 12-5i \end{array}$$

### Question 4

• Simplify and write your answer in the form of $a+ib$

(i) $\frac{-2}{1+i}$
(ii) $\frac{2+3i}{4-i}$
(iii) $\frac{9-7i}{3+i}$
(iv) $\frac{2-6i}{3+i}-\frac{4+i}{3+i}$
(v) $({1+i}/{1-i})^2$
(vi) $\frac{1}{(2+3i)(1-i)}$

Solution

(i) $$\begin{array}{cl} \frac{-2}{1+i} &= \frac{-2}{1+i}\times\frac{1-i}{1-i}\\ &= \frac{-2(1-i)}{(1+i)(1-i)}\\ &= \frac{-2+2i}{1-i^2}\\ &= \frac{-2+2i}{1+1}\\ &= \frac{-2+2i}{2}\\ &= -1+i \end{array}$$

(ii) $$\begin{array}{cl} \frac{2+3i}{4-i} &= \frac{2+3i}{4-i}\times\frac{4+i}{4+i}\\ &= \frac{2(4+i)+3i(4+i)}{(4-i)(4+i)}\\ &= \frac{8+2i+12i+3i^2}{16-i^2}\\ &= \frac{8+14i-3}{16+1}\\ &= \frac{5+14i}{17}\\ &= \frac{5}{17}+\frac{14i}{17} \end{array}$$

(iii) $$\begin{array}{cl} \frac{9-7i}{3+i}\\ &= \frac{9-7i}{3+i}\times\frac{3-i}{3-i}\\ &= \frac{9(3-i)-7i(3-i)}{9-(i^2)}\\ &= \frac{27-9i-21i+7i^2}{9+1}\\ &= \frac{27-30i-7}{10}\\ &= \frac{20-30i}{10} \end{array}$$

(iv) $$\begin{array}{cl} \frac{2-6i}{3+i}-\frac{4+i}{3+i}\\ &= \frac{{2-6i}-{(4+i)}}{3+i}\\ &= \frac{2-6i-4-i}{3+i}\\ &= \frac{-2-7i}{3+i}\\ &= \frac{-2-7i}{3+i}\times\frac{3-i}{3-i}\\ &= \frac{-2(3-i)-7i(3-i)}{9-i^2}\\ &= \frac{-6+2i-21i+7i^2}{9+1}\\ &= \frac{-6-19i-7}{10}\\ &= \frac{-13-19i}{10}\\ &= \frac{-13}{10}-\frac{19i}{10}\\ &= 2-3i \end{array}$$

(v) $$\begin{array}{cl} (\frac{1+i}{1-i})^2\\ &= \frac{1+i}{1-i}\times\frac{1+i}{1+i}^2\\ &= (\frac{1(1+i)+i(1+i)}{1-i^2})^2\\ &= (\frac{1+i+i+i^2}{1+1})^2\\ &= (\frac{2i}{2})^2\\ &= (i)^2\\ &= -1 \end{array}$$

(Vi) $$\begin{array}{cl} \frac{1}{(2+3i)(1-i)}\\ &= \frac{2-3i}{4-9i^2}\times\frac{1+i}{1-i^2}\\ &= \frac{2-3i}{4+9}\times\frac{1+i}{1+1}\\ &= \frac{2(1+i)-3i(1+i)}{13\times2}\\ &= \frac{2+2i-3i-3i^2}{26}\\ &= \frac{2-i+3}{26}\\ &= \frac{5-i}{26}\\ &= \frac{5}{26}-\frac{i}{26} \end{array}$$

### Question 5

Calculate (a) $\overline{z}$ (b) $z + \overline{z}$ © $z - \overline{z}$ (d) $z\overline{z}$ , for each of the following. (i) $z = -i$
(ii) $z = 2 + i$
(iii) $\frac{1+i}{1-i}$
(iv) $\frac{4-3i}{2+4i}$

Solution

(5)i $$\begin{array}{cl} z = -i\\ * (a) z = 0 - i\\ \overline{z} = 0 + i\\ * (b) z + \overline{z}\\ &=-i +i\\ &= 0\\ * (c) z - \overline{z}\\ &= -i-(i)\\ &= -i-i\\ &= -2i\\ * (d) z\overline{z}\\ &= (-i)(i)\\ &= -i^2\\ &= -(-1)\\ &= 1 \end{array}$$

5(ii) $$\begin{array}{cl} z = 2 + i\\ * (a) \overline{z} = 2 - i\\ * (b) z+ \overline{z}\\ & = (2+i)+(2-i)\\ & = 2+i+2-i\\ & = 4\\ * (c) z-\overline{z}\\ &= (2+i)-(2-i)\\ &= 2 +i-2+i\\ &= 2i\\ * (d) z\overline{z}\\ &= (2+i)(2-i)\\ &= 4-i^2\\ &= 4-(-1)\\ &= 4+1\\ &= 5 \end{array}$$

5(iii) $$\begin{array}{cl}\\ z =\frac{1+i}{1-i}\\ &= \frac{1+i}{1-i}\cdot\frac{1+i}{1+i}\\ &= \frac{(1+i)^2}{(1-i^2}\\ &= \frac{1+i^2+2i}{1+1}\\ &= \frac{1-1+2i}{2}\\ &= \frac{2i}{2}\\ &= i\\ * (a) \overline{z} = -i\\ * (b) z + \overline{z}\\ &= i-i\\ &= 0\\ * (c) z- \overline{z}\\ &= i-(-i)\\ &= i+i\\ &= 2i\\ *(d)z\overline{z}\\ &= (i)(-i)\\ &= -i^2\\ &= -(-1)\\ &= 1 \end{array}$$

5(iv) $$\begin{array}{cl} z = \frac{4-3i}{2+4i}\\ &=\frac{4-3i}{2+4i}\times\frac{2-4i}{2-4i}\\ &=\frac{(4-3i)(2-4i)}{2^2-(4i)^2}\\ &=\frac{4(2-4i)-3i(2-4i)}{4-16i^2}\\ &=\frac{8-16i-6i+12i^2}{4-16(-1)}\\ &=\frac{8-22i+12(-1)}{4+16}\\ &=\frac{-4-22i}{20}\\ &=\frac{-4}{20}-\frac{22i}{20}\\ &=\frac{-1}{5}-\frac{11i}{10}\\ * (a) \overline{z} = \frac{-1}{5} +\frac{11i}{10}\\ * (b) z + \overline{z}\\ &= (\frac{-1}{5} - \frac{11i}{10}) + (\frac{-1}{5} - \frac{11i}{10})\\ &= \frac{-1}{5} - \frac{11i}{10} - \frac{-1}{5} + \frac{11i}{10}\\ &= \frac{-1}{5} - \frac{1}{5}\\ &= \frac{-2}{5}\\ * (c) z - \overline{z}\\ &= (\frac{-1}{5} -\frac{11i}{10}) - (\frac{-1}{5} - \frac{11i}{10})\\ &= \frac{-1}{5} - \frac{11i}{10} + \frac{1}{5} -\frac{11i}{10}\\ &= \frac{-11i}{10} - \frac{11i}{10}\\ &= \frac{-22}{10}\\ &= \frac{-11i}{5}\\ * (d) z\overline{z}\\ &= (\frac{-1}{5} - \frac{11i}{10})(\frac{-1}{5} + \frac{11i}{10})\\ &= (\frac{-1}{5})^2 -(\frac{11i}{10})^2\\ &= \frac{1}{25} - \frac{121i^2}{100}\\ &= \frac{1}{25} + \frac{121}{100}\\ &= \frac{4+121}{100}\\ &= \frac{125}{100}\\ &= \frac{5}{4} \end{array}$$

### Question 6

If $z = 2 + 3i , w = 5 - 4i$, show that

• (i) $\overline{z+w} = \overline{z} + \overline{w}$
• (ii) $\overline{z-w} = \overline{z} -\overline{w}$
• (iii) $\overline{zw} = \overline{z}\overline{w}$
• (iv) $\overline{\frac{z}{w}} = \frac{\overline{z}}{\overline{w}}$
• (v) $\frac{1}{2}(z+\overline{z})$ is real part of z
• (vi) $\frac{1}{2i}(z-\overline{z})$ is the imaginary part of z

Solution
6(i) $$\begin{array}{cl} z = 2+3i\\ w = 5-4i\\ * (i) \overline{z+w} = \overline{z} + \overline{w}\\ \overline{z} = 2 -3i\\ \overline{w} = 5+4i\\ \overline{z}+\overline{w} = 2-3i+5+4i\\ &= 7+i\\ z + w = 2 + 3i+5-4i\\ &= 7-i\\ \overline{z+w} = 7+i\\ L.H.S =R.H.S \end{array}$$ 6(ii) $$\begin{array}{cl} \overline{z-w} = \overline{z} - \overline{w}\\ z-w = (2+3i)-(5-4i)\\ &= 2 +3i-5+4i\\ &= -3+7i\\ \overline{z-w} = -3-7i\\ \overline{z}-\overline{w} = (2-3i)-(5+4i)\\ &= 2-3i-5-4i\\ &= -3+7i\\ L.H.S = R.H.S \end{array}$$ 6(iii) $$\begin{array}{cl} \overline{zw} = \overline{z}\overline{w}\\ zw = (2+3i)(5-4i)\\ &= 2(5-4i)+3i(5-4i)\\ &= 10-8i+15i-12i^2\\ &= 10+7i-12(-1)\\ &= 10+12+7i\\ &= 22+7i\\ \overline{zw} = 22-7i\\ \overline{z}\overline{w} &= (2-3i)(5+4i)\\ &= 2(5+4i)-3i(5+4i)\\ &= 10+8i-15i-12i^2\\ &= 10-7i-12(-1)\\ &= 10+12-7i\\ &= 22-7i\\ L.H.S = R. H.S \end{array}$$ 6(iv) $$\begin{array}{cl} \overline{\frac{z}{w}} = \frac{\overline{z}}{\overline{w}}\\ \frac{z}{w} = \frac{2+3i}{5-4i}\\ &= \frac{2+3i}{5-4i}\times\frac{5+4i}{5+4i}\\ &= \frac{2(5+4i)+3i(5+4i)}{25-16i^2}\\ &= \frac{10+8i+15i+12i^2}{25-16(-1)}\\ &= \frac{10+23i+12(-1)}{25+16}\\ &= \frac{10-12+23i}{41}\\ &= \frac{-2-23i}{41}\\ \frac{\overline{z}}{\overline{w}} = \frac{2-3i}{5+4i}\\ &= \frac{2-3i}{5+4i}\times\frac{5-4i}{5-4i}\\ &= \frac{2(5-4i)-3i(5-4i)}{25-16i^2}\\ &= \frac{10-8i-15i+12i^2}{25-16(-1)}\\ &= \frac{10-23i+12(-1)}{25+16}\\ &= \frac{-2-23i}{41}\\ L.H.S = R.H.S \end{array}$$ 6(V) $$\begin{array}{cl} \frac{1}{2}(z+\overline{z}) &= \frac{1}{2}(2+3i+2-3i)\\ &= \frac{1}{2}(4)\\ &= 2 \hbox{ (it is real part of } z). \end{array}$$ 6(vi) $$\begin{array}{cl} \frac{1}{2i}(z-\overline{z}) &= \frac{1}{2}(2+3i-(2-3i))\\ &= \frac{1}{2}(2+3i-2+3i)\\ &= \frac{1}{2}(6i)\\ &= 3i \hbox{ (it is imaginary part of } z). \end{array}$$

### Question 7

Solve the following equations for real x and y

• (i) $(2-3i)(x+yi) = 4+i$
• (ii) $(3-2i)(x+yi)= 2(x-2yi)+2i-1$
• (iii) $(3+4i)^2-2(x-yi) = x+yi$

Solution
7(i) $$\begin{array}{cl} (2-3i)(x+yi) = 4+i\\ 2(x+yi)-3i(x+yi) = 4+i\\ 2x+2yi-3xi-3yi^2 = 4+i\\ 2x+2yi-3yi-3y(-1) = 4+i\\ (2x+3y)+(2y-3x)i = 4+i\\ 2x+3y =4 (i)\\ 2y-3x = 1 (ii)\\ 3\times(i) + 2\times(ii)\\ 6x+9y =12 (iii)\\ -6x+4y= 2 (iv)\\ 13y = 14\\ y = 14/13\\ \hbox{put value of y in (i)}\\ 2x+3(14/13) =4\\ \hbox{multiply 13 in both sides}\\ 2\times13x + 42/13\times13 = 52\\ 26x+42 = 52\\ 26x = 52-42\\ 26x = 10\\ x = 10/26\\ x = 5/13\\ * x = 5/13 , y = 14/13 \end{array}$$

7(ii) $$\begin{array}{cl} (3-2i)(x+yi)= 2(x-2yi)+2i-1\\ 3(x+yi)-2i(x+yi) = 2x-4yi+2i-1\\ 3x+3yi-2xi-2yi^2 = 2x-1+(2-4y)i\\ 3x+(3y-2x)i-2y(-1) = 2x-1 +(2-4y)i\\ (3x+2y) +(3y-2x)i = (2x-1)+(2-4y)i\\ 3x+2y = 2x-1\\ 3x-2x+2y = -1\\ x+2y = -1 (i)\\ 3y-2x = 2-4y\\ -2x+3y+4y = 2\\ -2x+7y = 2 (ii)\\ 2(i) + (ii)\\ 2x+4y = -2\\ -2x+7y = 2\\ 11y =0\\ y=0\\ \hbox{put value of y in (i)}\\ x+2(0) = -1\\ x = -1\\ * x = -1 ,y =0 \end{array}$$

7(iii) $$\begin{array}{cl} (3+4i)^2-2(x-yi) = x+yi\\ 3^2+24i+16i^2-2x+2yi = x + yi\\ 9+16(-1)-2x+24i+2yi = x+yi\\ 9-16-2x+(24+2y)i = x+yi\\ (-7-2x)+(24+2y)i = x +yi\\ -7-2x = x\\ -2x-x = 7\\ -3x = 7\\ x = -7/3\\ 24+2y = y\\ 2y-y = -24\\ y = -24\\ * x =-7/3 , y = -24 \end{array}$$

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