# Exercise 1.2 (Solutions)

Notes (Solutions) of Exercise 1.2: Textbook of Algebra and Trigonometry Class XI (Mathematics FSc Part 1 or HSSC-I), Punjab Textbook Board (PTB) Lahore.

The main topics of this exercise are complex numbers, real part and imaginary part of complex numbers, properties of the fundamental operation on complex numbers, complex number as ordered pair of real numbers and special subset of complex numbers. These notes are based on the new Student Learning Outcomes (SLOs). Version: 4.0, Available at MathCity.org

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

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

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

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

• fsc-part1-ptb/sol/ch01/ex1-2