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Question 1

Complex plane

Answer 1

ℂ = {z : x+iy with x & y ∈ ℝ}

ℝ subset of ℂ as any real x = x+0i is also complex

Although ℝ is ordered, ℂ is not.

(An ordered set is st x less than y , equal to or bigger than)

Therefore we can’t define inequalities for complex numbers

Question 2

When can we define inequalities involving complex numbers

Answer 2

When it involves the real part or the modulus of the complex number

Like |Z ₁| < |Z₂|. Or Re Z ₁ < ReZ₂

As modulus is related to distance we can see that one complex number is closer to the origin on the Argand diagram than the other

Or |z-a| > |z-b|. Z is closer to b than a etc

Question 3

Dealing with complex numbers:

Answer 3

• we can equate real and imaginary parts
• we can multiply fractions by the complex conjugate
• use the exp form

Question 4

Complex conjugate

Answer 4

Z bar

= x-iy

Is the complex conjugate of z =x +iy

Question 5

Modulus of z

|zw|

Answer 5

|Z| = sqrt( x ² + y ²)
This is not real part!!

Re( x +iy) = x

Im (x+iy) =y (NOT iy)

Modulus of z is the distance between z and the origin
|z-a| is the distance betweeen z and a

|zw|= |z||w|

Question 6

|z| ²

Answer 6

|z| ² = z x zbar

= ( x ² + y ²)

Which is useful as moduli are hard to manipulate

Eg
1/z. =. 1/(z • zbar) • (zbar). = (x-iy)/ (x ² + y ²)

(We have multiplied by the complex conjugate)

Question 7

|z| ² is equal to z ²?

Answer 7

|z| ² is NOT EQUAL to z ²

z^2 = ( x ² -y ²) +2ixy

They are only equal when z was a real number

Question 8

Argz

Answer 8

Z is not equal to 0

And line OZ makes angle theta with the real axis—— this is arg z

Arg z = θ , θ + 2 π , θ + 4 π,

….,

θ +/- 2n π

Integer multiples added

Arg z has infinitely many values and it not a function
Check arg values in both cos and sin
Tan^-1 y/x

Question 9

Polar form

Modulus argument form

Answer 9

z= x+iy = r exp ( i θ)

= r( cosθ + isin θ)

r=|z| is bigger than or equal to 0 and θ is any value of arg z

We use
Exp( i α) = cos α +i sin α

Exp( i (θ +2n π))
= cos ( θ +2n π) + I sin (θ +2n π)
= cos θ +isin θ

= exp( i θ )

Question 10

Multiplying and dividing complex numbers

Answer 10

We can use modulus argument form to help divide and multiply complex numbers without looking at real and imaginary part separately

Z= r( cos θ +i sin θ )

W = s ( cos ϕ + i sin ϕ)

z•w = r•s • ( cos (θ +ϕ) + i sin ( θ+ ϕ))
And if s is not equal to 0
z/w = (r/s) • ( cos ( θ -ϕ) + i sin( θ -ϕ))

Useful for powers

Question 11

De moivres theorem

Answer 11

Let θ ∈ ℝ and n ∈ N

Then

( cos θ + isin θ) ^n
= cos n θ + isin θ

( cos θ + isin θ) ^1/n has n different values

= cos ((θ/n)+ (2k π/n)) + isin ((θ/n)+ (2k π/n))

k= 0,1,2,… ,(n-1)

N values (including k=0)

NOTE THIS IS FOR
MODULUS =1

In terms of (exp {iθ} )^ 1/n

Has precisely n different values by
Exp{i((θ/n) + (2k π/n))

K= 0,1,2,.. , (n-1)!!!

• cos( theta/n) is always a riot

Question 12

De moivres theorem example

The 5 values of 1^ (1/5)
= (cos(0) +isin0)^1/5

Answer 12

De moivres theorem example

The 5 values of 1^ (1/5)
= (cos(0) +isin0)^1/5

(|z|= 1 and arg z = 0 +2n π) for n in integers

• cos(0/5) + isin(0/5)=1
• cos( 2 π/5) + isin( 2 π/5)
• cos ( 4 π/5) +isin(4 π/5)
• cos(6 π/5) +isin(6 π/5)
• cos(8 π/5) + isin(8 π/5)

(Adding 2 π/5 to each angle)

Question 13

Roots of a complex number

Answer 13

Cos( theta/n) is always a root

Nth roots of a complex number r(cos θ +isin θ ) st r is bigger than 0

• nth roots are root_n{ r} •[cos( (θ /n)+ (2k π/n)) + isin ( ((θ /n)+ (2k π/n))

K=0,…,n-1

Question 14

Example: Roos of z^3 + 8 =0

Answer 14

Z =(-8)^ (1/3)

(-8 = 8 cos( π))
Which are:

• 2( cos( π/3) + isin( (π/3))
• 2(cos( π) + isin( π))
• 2(cos( 5π/3) +isin(5π/3))

Question 15

Example

Sqrt(3) + i ) /(1-i

Answer 15

(Sqrt(3) + i ) /(1-i)

(Sqrt(3) + i) (1+i)
__________________
1+1
By complex conjugates

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

EASIER IN MODULUS ARGUMENT FORM

= (2exp( i π/6)) /(root(2) • exp ( -i π/4))
= sqrt(2) ( exp( i5 π/12))

= sqrt(2) ( cos ( 5 π/12) + i sin (5 π/12))

Question 16

Example find ( sqrt(3) + i) ^48

Answer 16

( sqrt(3) + i) ^48 = (2cos( π/6) + 2i sin (π/6)) ^48

= 2^48 • ( cos( 8 π) + i sin (8 π)) by de Moivres

USE modulus argument form when taking powers:

( 2exp( πi/6))^48
= 2^48 ( exp( 8πi)) = 2^48

Question 17

Example find the roots of( sqrt(3) + i) ^1/48

Answer 17

=(2e^ iπ/6) ^1/48
so one value is 2^(1/48) e^(iπ/288)

Write α = 2^(1/48) e^(iπ/288)i, say and let
ω = e^(2πi/48).

Then α, αω, αω2
,. . . ,αω47 are the 48 values of the 48th roots of √3 + i.

(pick a α then 47 others are these)

Question 18

Example

Find

MOD{ (a-b) /(1-abarb)} when a and b in C, a not equal to b and |a| =1

Answer 18

MOD{ (a-b) /(1-abarb)} when a and b in C, a not equal to b and |a| =1

We use a•abar =1

MOD( (a-b)/(1- (b/a)))
= MOD ( (( a-b)/(a-b) )• (1/(1/a)))
= MOD(a) = 1

Avoid Using real and imaginary parts if possible

Question 19

Inequalities between the real and imaginary pets of a complex number

Answer 19

|Re z| ≤ |Re z | + |Im z |

And

Since |x| ≤ ( x

Im z| ≤ |z| ≤ |Re z| + | Im z|

Question 20

Theorem 1.1 the triangle inequalities

Answer 20

The triangle inequalities state that if z and w in the complex plane then

|| z| - |w|| ≤ |z-w| ≤ |z| + |w|

|z| - |w| | ≤ | z +w | ≤ |z| + |w|

Question 21

Example: show that

(1/3) ≤
|(2z²-1)/(z+2)|
≤ 3

for all z on |z|=1

Answer 21

|2z² − 1| ≤ 2|z²| + 1 = 2 + 1 = 3

|2z² − 1| ≥ | |2z²| − 1 | = 2 − 1 = 1
for all z such that |z| = 1.

1 = 2 − 1 = | |z| − 2| ≤ |z + 2| ≤ |z| + 2 = 1 + 2 = 3.

for all complex numbers z such that |z| = 1.

Question 22

Example: Show that
2 ≤ | 3z + 4i| ≤ 8
for all z with | z + 1| ≤ 1.

Answer 22

|3z+4i| = | 3(z+1) -3+4i|
|3(z+1) + |-3+4i|
= 3|z+1| + 5 ≤ 3+5 = 8

|3z+4i| = | 3(z+1) -3+4i|
≥ | |3(z + 1)| − |4i − 3| | = 5 − 3|z + 1| ≥ 5 − 3 = 2 ,

Question 23

Theorem 1.1 the triangle inequalities proof

The triangle inequalities state that if z and w in the complex plane then

|| z| - |w|| ≤ |z-w| ≤ |z| + |w|

|z| - |w| | ≤ | z +w | ≤ |z| + |w|

Answer 23

Proof
consider 
|z+w|² = ( z+w) (̅ ̅z ̅+ ̅w ̅)̅
= (z+w)( z̅  + w̅)
=zz̅  + ( zw̅ + z̅w) + ww̅
=zz̅  +2Re (zw̅) + ww̅
≤ |z|² + 2|zw̅| + |w|²
=|z|² + 2|z|w| + |w|²
= ( |z| + |w| )²

taking positive square roots to obtain |z+w| ≤ |z| + |w|
So
|z| = |(z − w) + w| ≤ |z − w| + |w|,

rearranging and interchanging
|z − w| = |w − z| ≥ |w| − |z|

( and replacing w with -w: |z − w| ≤ |z| + | − w| = |z| + |w|)

Combining |z − w| ≥ | |z| − |w||
finally replacing w with -w:
|z + w| ≥ | |z| − |w||

Question 24

Re(z)

Im(z)

Answer 24

Re(z) = (z + ̅z ) /2

Im(z) = (z - ̅z ) /2i

Question 25

|re^iθ|

Answer 25

|re^iθ| = |r| |e^iθ| = |r|

Question 26

(̅ ̅z ̅+ ̅w ̅)̅

(̅ ̅z ̅w ̅)̅

Answer 26

(̅ ̅z ̅+ ̅w ̅)̅ = ̅z + ̅w

(̅ ̅z ̅w ̅)̅ = ̅z ̅w

Question 27

EXAMPLE: use 4th roots of -1 to express z^4 +1 as a product of 2 real quadratics

Answer 27

x^4 +1 = (x^2 - √2 x + 1)(x^2 + √2 x +1)

Question 28

e^iθ + e^-iθ

Answer 28

e^iθ + e^-iθ = 2cosθ

Question 29

e^iθ - e^-iθ

Answer 29

e^iθ - e^-iθ = 2isinθ

Question 30

z^4 =1 EQUIV

Answer 30

z^4 =1

z^2 - 1)(z^2+1

Question 31

(1-z ^(k+1))

Answer 31

(1-z ^(k+1))

= (1-z)(1+z+z^2+…+z^k)

Question 32

e^iθ
=1
=-1
=i

Answer 32

e^iθ
=1 θ=2kπ
=-1 θ=(2k+1)π
=i θ=(1/2)(2k+1)π