Comple Number Exercise 6.1 | Basic Mathematics Solution [NEB UPDATED]

Exercise 6.1

1) Evaluate: 

Solution:

a)      (1,0)² = 12 = 1

b)      (1,0)³ = 13 = 1

c)       (0,1)⁵ = i5 = (i2)2.i = (-1)2.i = 1.i = i

d)      (0,i)¹¹ = i11 = (i2)5.i = -1.-I = -i.

2) Find the values of x and y in each of the following:

Solution:

a) (x,y) = (2 + 3,3 + 2) = (5,5). So, x = 5 , y = 5.

b) (x,y) = (2 – 2,1 – 1) = (0,0). So, x = 0 , y = 0.

c) (x,y) = (2,3) – (3,2) = (2 – 3,3 – 2) = (-1,1). So, x = -1 , y = 1.

d) (2,3) = (1,1) + (x + y) = (1 + x,1 + y)

Here, 2 = 1 + x, 3 = 1 + y

So, x = 1 and y = 2.

e) (x,y) = (1,1).(2,3)

(1 + i).(2 + 3i) = 2 + 3i + 2i + 3i² = - 1 + 5i

So, x = -1 and y = 5.

f) (x.y) = $\frac{{\left( {1,1} \right)}}{{\left( {3,4} \right)}}{\rm{\: }}$

Or, x + iy = $\frac{{\left( {1+i} \right)}}{{\left( {3+ 4i} \right)}}{\rm{\: }}$ * $\frac{{\left( {3-4i} \right)}}{{\left( {3-4i} \right)}}{\rm{\: }}$

= $\frac{{\left( {7- i} \right)}}{{\left( {9+16} \right)}}{\rm{\: }}$

 = $\frac{{\left( {7- i} \right)}}{{\left( 25 \right)}}{\rm{\: }}$                                

= 7/25 + i(-1/25)

So, x =  7/25, y = -1/25

 

3) Simplify: 

a) $\sqrt{ - 9} $ + $\sqrt { - 25} $ – $\sqrt { - 36} $

Solution:

$\sqrt{ - 9} $ + $\sqrt { - 25} $ – $\sqrt { - 36} $

Or, $\sqrt{ - 9} $ + $\sqrt { - 25} $ – $\sqrt { - 36} $

= $\sqrt { - 1{\rm{*}}9} $ + $\sqrt { - 1{\rm{*}}25} $ – $\sqrt { - 1{\rm{*}}36} $.

= $\sqrt {{{\rm{i}}^2}{\rm{*}}{3^2}} $ + $\sqrt {{{\rm{i}}^2}{\rm{*}}{5^2}} $ – $\sqrt {{{\rm{i}}^2}{\rm{*}}{6^2}} $

= 3i + 5i – 6i = 2i. 

b) $\left( {3 - \sqrt { - 4} } \right)\left( {2 + \sqrt { - 9} } \right)$

Solution:

$\left( {3 - \sqrt { - 4} } \right)\left( {2 + \sqrt { - 9} } \right)$

Or, $\left( {3 - \sqrt { - 4} } \right)\left( {2 + \sqrt { - 9} } \right)$ = $\left( {3 - \sqrt { - 1{\rm{*}}4} } \right) + \left( {2 + \sqrt { - 1{\rm{*}}9} } \right)$.

= $\left( {3 - \sqrt {{{\rm{i}}^2}{\rm{*}}{2^2}} } \right)\left( {2 + \sqrt {{{\rm{i}}^2}{\rm{*}}{3^2}} } \right)$ = (3 – 2i)(2 + 3i) = 6 + 9i – 4i – 6i².

= 6 + 5i – 6(-1) = 6 + 5i + 6 = 12 + 5i. 

c) $\sqrt{ - 16} $.$\sqrt { - 1} $

Solution:

$\sqrt{ - 16} $.$\sqrt { - 1} $

Or, $\sqrt{ - 16} $.$\sqrt { - 1} $

= $\sqrt { - 1{\rm{*}}16} $.$\sqrt { - 1} $

= $\sqrt {{{\rm{i}}^2}{\rm{*}}{4^2}} .\sqrt {{{\rm{i}}^2}} $

= 4i.i

= 4i²

= 4(-1)

= -4 

d) 3i² + i³ + 9i⁴ – i⁷

Solution:

3i² + i³ + 9i⁴ – i⁷

= 3(-1) + i².i + 9(i²)² – (i²)³.i = -3 + (-1).i + 9(-1)² – (-1)³.i

= -3 – I + 9.1 – (-1).i = -3 – I + 9 + i = 6. 

e) $\frac{1}{{\rm{i}}} - \frac{1}{{{{\rm{i}}^2}}} + \frac{1}{{{{\rm{i}}^3}}} - \frac{1}{{{{\rm{i}}^4}}}

Solution:

$\frac{1}{{\rm{i}}} - \frac{1}{{{{\rm{i}}^2}}} + \frac{1}{{{{\rm{i}}^3}}} - \frac{1}{{{{\rm{i}}^4}}}

Or, $\frac{1}{{\rm{i}}} - \frac{1}{{{{\rm{i}}^2}}} + \frac{1}{{{{\rm{i}}^3}}} - \frac{1}{{{{\rm{i}}^4}}} = \frac{1}{{\rm{i}}} - \frac{1}{{ - 1}} + \frac{1}{{{{\rm{i}}^2}.{\rm{i}}}} - \frac{1}{{{{\left( {{{\rm{i}}^2}} \right)}^2}}}$.

= $\frac{1}{{\rm{i}}} + 1 - \frac{1}{{\rm{i}}} - \frac{1}{{{{\left( { - 1} \right)}^2}}}$

= $\frac{1}{{\rm{i}}} + 1 - \frac{1}{{\rm{i}}} - 1$ = 0 

f) $\frac{1}{{\rm{i}}} + \frac{1}{{{{\rm{i}}^2}}} + \frac{1}{{{{\rm{i}}^3}}} + \frac{1}{{{{\rm{i}}^4}}}$

Solution:

Or, $\frac{1}{{\rm{i}}} + \frac{1}{{{{\rm{i}}^2}}} + \frac{1}{{{{\rm{i}}^3}}} + \frac{1}{{{{\rm{i}}^4}}}$

= $\frac{1}{{\rm{i}}} + \frac{1}{{ - 1}} + \frac{1}{{{{\rm{i}}^2}.{\rm{i}}}} + \frac{1}{{{{\left( {{{\rm{i}}^2}} \right)}^2}}}$

= $\frac{1}{{\rm{i}}} - 1 + \frac{1}{{ - 1.{\rm{i}}}} + \frac{1}{{{{\left( { - 1} \right)}^2}}}$

= $\frac{1}{{\rm{i}}} - 1 - \frac{1}{{\rm{i}}} + 1$

= 0.

 

4) Prove that:

a) (1 + i)⁴${\left( {1 + \frac{1}{{\rm{i}}}} \right)^4}$=16

Solution:

LHS = (1 + i)⁴${\left( {1 + \frac{1}{{\rm{i}}}} \right)^4}$= (1 + i)⁴$\frac{{{{\left( {{\rm{i}} + 1} \right)}^4}}}{{{{\rm{i}}^4}}}{\rm{\: }}$                           

= $\frac{{{{\left( {1 + {\rm{i}}} \right)}^8}}}{{{{\left( {{{\rm{i}}^2}} \right)}^2}}}$ = $\frac{{{{\left[ {{{\left( {1 + {\rm{i}}} \right)}^2}} \right]}^4}}}{{{{\left( { - 1} \right)}^2}}}{\rm{\: }}$= $\frac{{{{\left( {1 + 2{\rm{i}} + {{\rm{i}}^2}} \right)}^4}}}{1}$

= $\frac{{{{\left( {1 + 2{\rm{i}} - 1} \right)}^4}}}{1}$ = (2i)⁴ = 16i⁴ = 16.(i²)² = 16.(-1)² = 16 = R.H.S.

b) ${(1 - {i^3})^6}{\left( {1 - \frac{1}{{{i^3}}}} \right)^6} = 64$

Solution:

 

5) Find the real value of x and y if:

a) x + iy = (2-3i)(3-2i)

Solution:

(x + iy) = (2 – 3i)(3 – 2i) = 6 – 4i – 9i + 6i² = 6 – 13i – 6 = -13i

X + iy = 0 – 13i

So,x = 0 and y= -13. 

b) (x-1)i + (yy+1) = (1+i)(4-3i)

Solution:

(x – 1)I + (y + 1) = (1 + i)(4 – 3i)

(y + 1) + (x – 1)I = 4 – 3i + 4i – 3i²

(y + 1) + (x – 1)I = 4 + I – 3(-1)

(y + 1) + (x – 1)I = 4 + I + 3.

(y + 1) + (x – 1)I = 7 + i

Hence, y + 1 = 7 and x – 1 = 1

So, x = 2 and y = 6.

 

6) Show that $\frac{{3 + 2i}}{{2 - 5i}} + \frac{{3 - 2i}}{{2 + 5i}}$ is purely a number.

Solution:

Or, $\frac{{\left( {3 + 2{\rm{i}}} \right)\left( {2 + 5{\rm{i}}} \right) + \left( {3 - 2{\rm{i}}} \right)\left( {2 - 5{\rm{i}}} \right)}}{{\left( {2 - 5{\rm{i}}} \right)\left( {2 + 5{\rm{i}}} \right)}}$

= 6 + 15i + 4i + 10i² + 6 – 15i – 4i + 10i²

= 12 + 20i²

= 12 + 20(-1) = 12 – 20 = -8.

Hence, the given expression is purely a real number.