Solutions To Simplifying Complex Expressions & Solving Complex Equations (Algebra II Review)

1)  Write: (2 + 3i)/ (1 + 2i) in the form a + bi

Soln.

Multiply both numerator and denominator by the complex conjugate of denominator, i.e. (1 - 2i)

So take:  (2 + 3i)/ (1 + 2i)  [(1 - 2i)/ (1 - 2i)]

Expand numerator:

(2 + 3i) (1 - 2i) = 2(1) + 2(-2i) + 3i(1) + 3i (-2i)

= 2 - 4i + 3i - 6 i ² 

But  i ² =   -1 

Þ

  2 - i - 6 (- 1) = 2 - i + 6 =  8 - i

2) Show the product of a complex number a + bi and its conjugate (a - bi) is a real number. (You can use the result from (1).

Soln.

(8 - i) (8 + i)  =  8(8) + 8(i) - i(8) + i(-i) 

= 64 +8i - 8i -(-1) = 65

3)  Find the product: (3 - 5i) (2 + i)

Soln.

(3 - 5i) (2 + i)  = 3 (2) + 3(i) - 5i(2) + (-5i)(i)

=  6 + 3i - 10i  - 5((-1) =  11 - 7i

4)  Write in the form a + bi:

Soln.

1/5 (4) - 1/5 (2i) - i/10 (4) -  i/10 (2i) =

4/5 + 2i/5 - 4i/10 -2 (-1)/10= 0.8 + 0.4i  - 0.4i + 0.2 =  1

Solve each of the following:

5)   x⁵   +  32  = 0  for all roots  

Soln.

x⁵   =  - 32  + 0i = 32 cis ( 180^o     + n 360^o )

x =[32 cis (180^o     + n 360^o )]^1/5

Letting n = 0, 1, 2, 3, 4 in order:

x1 =  2 cis 36^o 

x2 = 2 cis 108^o 

x3 =  2 cis 180^o 

x4 =  2 cis 324^o 

6) x⁴   - 1  = 0

Soln.

x⁴  = -1  

This is just the 4th roots of unity, i.e.

x = (-1)^1/4

Let:  w_n = cos (2 p)k/ n + isin(2 pk)/n

For k = 0, 1, 2 and 3 then,

The first root::  w₀ = cos(0) + isin(0) = 1

The second root: w₁ = cos (p/2) + isin(p/2) = 1i = i

The third root: w₂ = cos (p) + isin(p) = -1

And the fourth root: w₃ = cos(3 p/2) + isin(3p/2) = -1i = -i

7) x²   - 4x  + 8 = 0

Soln.

Apply the quadratic formula:

x = -b + Ö {b² - 4ac}/ 2a

a = 1, b = -4, c  = 8

Discriminant:

b² - 4ac =  16 - 4 (1) (8) = 16 - 32 = -16

Note: both roots are complex if discriminant is negative

x =  4 + Ö {-16}/ 2  = 4  + {Ö4Ö4 Ö-1 }/ 2  

  Ö-1  =  i  

Then: x = 2 +  2i, 2 -  2i

8) x²   + 10x + 29 = 0

Soln.

Apply the quadratic formula:

x = -b + Ö {b² - 4ac}/ 2a

a = 1, b = 10, c  = 29

Discriminant:

b² - 4ac =  100 - 4 (1) (29) = 100 - 116 = -16

Negative so both roots complex

x =  -10 + Ö {-16}/ 2  = -10  + {Ö4Ö4 Ö-1 }/ 2

x = -5 -  2i, -5 +  2i

9) 4x²   - 4x  + 5 = 0

Soln.   Apply the quadratic formula:

x = -b + Ö {b² - 4ac}/ 2a

a = 4, b = -4, c  = 5

Discriminant:

b² - 4ac =  16 - 4 (-4) (5) = 16 - 80 = -64

Negative so both roots complex,

 x =  4 + Ö {-64}/ 2  = 4  + {Ö4Ö16 Ö-1 }/ 8

But: i = Ö-1 

x = -1/2 +  i, -1/2 -  i  = - 0.5 - i

10)4 x - 7  = x²  

Soln.  Re-arrange to get proper quadratic eqn.:

 x²  -  4x +  7 = 0

Apply the quadratic formula:

x = -b + Ö {b² - 4ac}/ 2a

a = 1, b = -4, c  = 7

Discriminant:

b² - 4ac =  16 - 4 (1) (7) = 16 - 28 = -12

Negative so both roots complex,

 x =  4 + Ö {-12}/ 2  = 4  + {Ö4Ö3 Ö-1 }/ 2

But i = Ö-1

Then: x = 2 + Ö3i, 2 - Ö3i