Exercise 1.2 (Solutions)

Question 1 Verify the addition properties of complex numbers.

Solution Let $z,w,$ and $v$ be complex numbers. Then the following properties hold.

- Commutative Law for Addition $$z+w=w+z\nonumber$$ - Additive Identity $$z+0=z\nonumber$$ - Existence of Additive Inverse $$\begin{array}{l} \mbox{For each} \; z\in \mathbb{C}, \mbox{there exists}\; -z\in \mathbb{C} \mbox{ such that}\; z+\left( -z\right) =0 \\ \mbox{In fact if } z=a+bi, \mbox{ then } -z=-a-bi. \end{array}\nonumber$$ - Associative Law for Addition $$\left( z+w\right) +v= z +\left( w+v\right)\nonumber$$

Question 2 Verify the multiplication properties of the complex numbers.

Question 3 Verify the distributive law of complex numbers.

Question 4(i)

Simplify: $i^9$

Solutions

$i^9$ $= (i^2)^4 \cdot i$ $=1 \cdot i$ =$i$.

Question 4(ii)

Simplify: $i^{14}$

Solutions

$i^{14}$ $=(i^2)^7$ $=(-1)^7$ $=-1$.

Question 4(iii)

Simplify: ${-i}^{19}$

Solutions

$$\begin{align} {-i}^{19}& =[(-1)(i)] ^{19}=(-1)^{19}\cdot i^{19} \\ & =-1\cdot i^{18}\cdot i = -(i^2)^{9} \cdot i \\ & =-(-1)^{9} i=-(-1)i =i. \end{align}$$

Question 4(iv)

Simplify: $\displaystyle {{(-1)}^{-\frac{21}{2}}}$

Solution $$\begin{align} (-1)^{-\frac{21}{2}}&=\frac{1}{(-1)^\frac{21}{2}}=\frac{1}{[(-1)^\frac{1}{2}]^{21}}\\ &= \frac{1}{i^{21}}=\frac{1}{(i^2)^{10}\cdot i} \\ &=\frac{1}{(-1)^{10}\cdot i}=\frac{1}{1 \cdot i}\\ &=\frac{1}{i}=\frac{1}{i}\times \frac{i}{i}\\ &=\frac{i}{i^2}=\frac{i}{-1}=-i. \end{align}$$

Question 5(i)

Write $\sqrt{-1}b$ in term of $i$

Solutions

$\sqrt{-1}b= bi$

Question 5(ii)

Write $\sqrt{-5}$ in term of $i$

Solutions

$\sqrt{-5}= \sqrt{(-1)5} =\sqrt{-1}\cdot \sqrt{5} =\sqrt{5}i$

Question 5(iii)

Write $\sqrt{-\frac{16}{25}}$ in term of $i$

Solutions

$\sqrt{-\frac{16}{25}}= \sqrt{(-1)\frac{16}{25}}$ $=\sqrt{-1}\cdot \sqrt{\frac{4^2}{5^2}} =\frac{4}{5}i$

Question 5(iv)

Write $\sqrt{\frac{1}{-4}}$ in term of $i$

Solutions

$\displaystyle \quad \sqrt{\frac{1}{-4}}= \sqrt{\frac{1}{(-1)4}} = \frac{1}{\sqrt{(-1)4}}$

$\displaystyle =\frac{1}{2i}= \frac{1}{2i}\times \frac{i}{i} =\frac{i}{2i^2}$

$\displaystyle =\frac{i}{2(-1)}=\frac{i}{-2}=-\frac{i}{2}$

Question 6

Simplify: $(7,9)+(3,-5)$

Solutions

$\quad (7,9)+(3,-5)$

$=(7+3,9-5)=(10,4)$

Question 7

Simplify: $(8,-5)-(-7,4)$

Solutions

$\quad (8,-5)-(-7,4)$

$=(8+7,-5-4)=(15,-9)$

Question 8

Simplify: $(2,6)\cdot(3,7)$

Solutions

$\quad (2,6)\cdot(3,7)$

$=(2 \cdot 3-6\cdot7,2\cdot7+6\cdot3)$

$=(6-42,14+18)=(-36,32)$

Question 9

Simplify: $(5,-4)\cdot(-3,-2)$

Solutions

$\quad (5,-4)\cdot(-3,-2)$

$=((5)(-3)-(-4)(-2),(5)(-2)+(-4)(-3))$

$=(-15-8,-10+12)=(-23,2)$

Question 10

Simplify: $(0,3)\cdot (0,5)$

Solutions

$\quad (0,3)\cdot (0,5)$

$=((0)(0)-(3)(5),(0)(5)+(3)(0))$

$=(0-15,0+0)=(-15,0)$

Question 11

Simplify: $(2,6)\div (3,7)$

Solutions

$\quad (2,6)\div (3,7)=\dfrac{(2,6)}{\left( 3,7 \right)}$

$=\dfrac{2+6i}{3+7i}=\dfrac{2+6i}{3+7i}\times \dfrac{3-7i}{3-7i}$

$=\dfrac{6-14i+18i-42{{i}^{2}}}{{{\left( 3 \right)}^{2}}-{{\left( 7i \right)}^{2}}}$

$=\dfrac{6+4i-42\left( -1 \right)}{9-49{{i}^{2}}}$

$=\dfrac{6+4i+42}{9-49\left( -1 \right)}=\dfrac{48+4i}{9+49}$

$=\dfrac{48+4i}{58} =\dfrac{48}{58}+\dfrac{4}{58}i$

$=\dfrac{24}{29}+\dfrac{2}{29}i=\left( \dfrac{24}{29},\dfrac{2}{29} \right)$

Question 12

Simplify: $(5,-4)\div (-3,-8)$

Solutions

$\quad (5,-4)\div (-3,-8)=\dfrac{(5,-4)}{\left( -3,-8 \right)}$

$=\dfrac{5-4i}{-3-8i}=\dfrac{5-4i}{-3-8i}\times \dfrac{-3+8i}{-3+8i}$

$=\dfrac{-15+40i+12i-32{{i}^{2}}}{{{\left( -3 \right)}^{2}}-{{\left( 8i \right)}^{2}}}$

$=\dfrac{-15+52i-32\left( -1 \right)}{9-64{{i}^{2}}}$

$=\dfrac{-15+52i+32}{9+64}$

$=\dfrac{17+52i}{73} =\dfrac{17}{73}+\frac{52}{73}i$

$=\left( \dfrac{17}{73},\dfrac{52}{73} \right)$.

Question 13

Prove that the sum as well as the product of any two conjugate complex number is a real number.

Solutions

Let the two conjugate complex number be $z =x+iy$ and $\overline{z }=x-iy$, where $x,y\in \mathbb{R}$

$$\begin{align} \text{Sum} &=z+\overline{z}\\ &=x+iy+x-iy \\ &=2x \in \mathbb{R}, \text{ as } x,y \in \mathbb{R}. \end{align}$$ Now $$\begin{align} \text{Product}&=z\cdot \overline{z} \\ &=\left( x+iy \right)\left( x-iy \right)\\ & ={{x}^{2}}-{{i}^{2}}{{y}^{2}} \\ &={{x}^{2}}-\left( -1 \right){{y}^{2}} \\ &={{x}^{2}}+{{y}^{2}} \in \mathbb{R}, \text{ as } x, y \in \mathbb{R}. \end{align}$$ Hence, we proved that sum as well as the product of any two conjugate complex number is a real number.

Question 14(i)

Find the multiplicative inverse of complex number $\left( -4,7 \right)$.

Solutions

Suppose $z=\left( -4,7 \right)$, then

multiplicative inverse of $z$ $=z^{-1} =\dfrac{1}{z}$

$=\dfrac{1}{\left( -4,7 \right)}=\dfrac{1}{-4+7i}$

$=\dfrac{1}{-4+7i}\times \dfrac{-4-7i}{-4-7i}$

$=\dfrac{-4-7i}{{{\left( -4 \right)}^{2}}-{{\left( 7i \right)}^{2}}}=\dfrac{-4-7i}{16-49(-1)}$

$=\dfrac{-4-7i}{16+49}=\dfrac{-4-7i}{65}$

$=\dfrac{-4}{65}-\dfrac{7i}{65}=\left( \dfrac{-4}{65},\frac{-7}{65} \right)$

Question 14(ii)

Find the multiplicative inverse of complex number $\left( \sqrt{2},-\sqrt{5} \right)$.

Solutions

Let $z=\left( \sqrt{2},-\sqrt{5} \right)$.

Multiplicative inverse of $z$ $=z^{-1}=\dfrac{1}{z}$

$=\dfrac{1}{\sqrt{2},-i\sqrt{5}}=\dfrac{1}{\sqrt{2}-i\sqrt{5}}$

$=\dfrac{1}{\sqrt{2}-i\sqrt{5}}\times \dfrac{\sqrt{2}+i\sqrt{5}}{\sqrt{2}+i\sqrt{5}}$

$=\dfrac{\sqrt{2}+i\sqrt{5}}{2-{{i}^{2}}\cdot 5}=\dfrac{\sqrt{2}+i\sqrt{5}}{2+5}$

$=\dfrac{\sqrt{2}+i\sqrt{5}}{7}=\dfrac{\sqrt{2}}{7}+\dfrac{\sqrt{5}}{7}i$

$=\left( \dfrac{\sqrt{2}}{7},\dfrac{\sqrt{5}}{7} \right)$

Question 14(iii)

Find the multiplicative inverse of complex numbers $\left( 1,0 \right)$.

Solutions

Let $z=\left( 1,0 \right)$.

Multiplicative inverse of $z$ $=z^{-1}=\dfrac{1}{z}$

$\dfrac{1}{\left( 1,0 \right)}=\dfrac{1}{1+0 i}=\dfrac{1}{1}=1$

$=1+0i=\left( 1,0 \right)$.

Question 15(i)

Factorize: ${{a}^{2}}+4{{b}^{2}}$

Solutions

$\quad {{a}^{2}}+4{{b}^{2}}={{a}^{2}}-\left( -1 \right)4{{b}^{2}}$

$={{a}^{2}}-{{i}^{2}}{{2}^{2}}{{b}^{2}} ={{\left( a \right)}^{2}}-{{\left( 2ib \right)}^{2}}$

$=\left( a+2ib \right)\left( a-2ib \right)$ —-

Question 15(ii)

Factorize: ${9{a}^{2}}+16{{b}^{2}}$

Solutions

$\quad {9{a}^{2}}+{16{b}^{2}}={{3a}^{2}}-\left( -1 \right)16{{b}^{2}}$

$={{3a}^{2}}-{{i}^{2}}{{4}^{2}}{{b}^{2}} ={{\left( 3a \right)}^{2}}-{{\left( 4ib \right)}^{2}}$

$=\left( 3a+4ib \right)\left( 3a-4ib \right)$

Question 15(iii)

Factorize: $3{{x}^{2}}+3{{y}^{2}}$

Solutions

$\quad 3{{x}^{2}}+3{{y}^{2}}=3\left[ {{x}^{2}}-\left( -1 \right){{y}^{2}} \right]$

$=3\left[ {{x}^{2}}-{{i}^{2}}{{y}^{2}} \right] =3\left[ {{\left( x \right)}^{2}}-{{\left( iy \right)}^{2}} \right]$

$=3\left( x+iy \right)\left( x-iy \right)$

Question 16(i)

Separate into real and imaginary parts: $\dfrac{2-7i}{4+5i}$ (write into simple complex number)

Solutions

$\quad \dfrac{2-7i}{4+5i}$

$=\dfrac{2-7i}{4+5i}\times \dfrac{4-5i}{4-5i}$

$=\dfrac{8-10i-28i+35{{i}^{2}}}{{{\left( 4 \right)}^{2}}-{{\left( 5i \right)}^{2}}}$

$=\dfrac{8-38i+35\left( -1 \right)}{16-25\left( -1 \right)}=\dfrac{8-38i-35}{16+25}$

$\dfrac{-27-38i}{41}=-\dfrac{27}{41}-\dfrac{38}{41}i$

Question 16(ii)

Separate into real and imaginary parts $\dfrac{{{\left( -2+3i \right)}^{2}}}{\left( 1+i \right)}$ (write into simple complex number)

Solutions

$\dfrac{{{\left( -2+3i \right)}^{2}}}{\left( 1+i \right)}=\dfrac{4+9{{i}^{2}}-12i}{1+i}$

$=\dfrac{4-9-12i}{1+i}=\dfrac{-5-12i}{1+i}$

$=\dfrac{-5-12i}{1+i}\times \dfrac{1-i}{1-i}$

$=\dfrac{-5+5i-12i+12{{i}^{2}}\left( -1 \right)}{1-{{i}^{2}}}$

$=\dfrac{-5-7i+12\left( -1 \right)}{1-\left( -1 \right)}$

$=\dfrac{-17-7i}{2}=-\dfrac{17}{2}-\dfrac{7}{2}i$

Question 16(iii)

Separate into real and imaginary parts: $\dfrac{i}{1+i}$ (write into simple complex number)

Solutions

$\quad \dfrac{i}{1+i}=\dfrac{i}{1+i}\times \dfrac{1-i}{1-i}$

$=\dfrac{i-{{i}^{2}}}{1-{{i}^{2}}}=\dfrac{i-\left( -1 \right)}{1-\left( -1 \right)}$

$=\dfrac{i+1}{2}=\dfrac{1+i}{2}=\dfrac{1}{2}+\dfrac{i}{2}$