Solutions To Complex Functions & Equations Revisited

The Problems again:

1)  (a) Given u(x,y) + iv(x,y) = 2x² + i2y²  

find f(z,z*)

(b)  Express f(z)= z² + z – 3 in polar form

2) Solve for x and y:   8x 2 + 3iy  -  4    = 8y   – 4iy

3) Solve: (z+1) 3  = z 3

Solutions:

(1)   Let:  z = x + iy, and z* = x – iy

Adding:   z + z* = (x + iy) + (x - iy) = 2x

We see: x = (z + z*)/2

Subtracting:

(z – z*) = [x + iy – x + iy] =  i2y

Then:  y = (z – z*)/ 2i

We can now formulate the function f(z,z*):

f(z,z*) = 2[(z + z*)/2]²  + i2[(z – z*)/ 2i]²

2)  z² = r² exp(i2(q)) = r² (cos (2q) + isin(2 q))

z = r exp(i(q)) = r(cos(q) + isin(q)

so: z² + z = r² (cos (2q) + isin(2 q)) +  r(cos(q) + isin(q)

Collecting like terms in i and simplifying:

f(z) =  r²(cos (2q) + r(cos(q)) + i{sin(2q) + sin(q)} – 3

so: iv(r, q) =  i{sin(2q) + sin(q)}

and v (r, q) =  {sin(2q) + sin(q)}

while:

u(r, q) =  r² (cos (2q) + r(cos(q)) – 3

3)     8x 2 + 3iy  -  4    = 8y   – 4iy

 8x 2    =   4

So:   x 2    =   4/ 8   =  1/ 2

Therefore:  x =  Ö  ( 1/2)  =   1/ Ö2

=  Ö 2  /  2

  

And:   

8 y   =   - 3iy  - 4iy   =   - 7 iy

y  =   -7 i/  8

4)  Expand the left side and set equal to the right:

z ³ + 3z ² +  3z + 1   =  z ³

è3z ² +3z + 1   = z ³    -    z ³
  
 or  3z ²  + 3z +1 =  0

(This can be solved using the quadratic formula, to give two roots)

Then:  z1 = ½ + i Ö (3) / 6 

And z2 = -(½ ) + iÖ (3)/ 6

Checking the result against the  original equation:

z ³  = 0.192i and  (z ³ + 1) ³  = 0.192i

So both quantities are equal, the roots are correct.

Complex Functions & Solving Complex Equations Revisited

Basically,  complex functions  are analogous to regular (real)  functions, e.g.

f(x) = 3x – 4

f(x) = 2x² - 3x + 1

Except that complex variables of the form z = x + iy are incorporated. All the basic operations that apply to complex algebra, including polar forms, e.g.  r = exp[(i q)] ,  apply also to complex functions. Note that since complex functions are dependent on the complex variable z, we typically write them as f(z).

A good way to get started is by applying simple operations to functions.

For example, let f(z) = e ^(-3z)

Find the real and imaginary parts of the function f(z)

Since z = x + iy, we may write:

f(z) = exp[-3(x + iy)] = exp(-3x) [exp(-i3y)]

and

f(z) = exp(-3x)[cos (3y) – isin(3y)]

Then :   Re f(z) =  Re  exp(-3z) = exp(-3x)cos (3y)

And :  Im f(z) = Im exp(-3z) =  -iexp(-3x) sin (3y)

Finding numerical values for complex functions in many ways resembles the way we do it for real functions, simply substitute the f-value to be found into the function f(z), viz.

 

Find f(2i) for f(z) =  - 3z²

è       f(2i) = -3(2i)² =  -3 (-4) = 12

More examples:

1. Find: f(-3i) for f(z) = (z + 2 – 3i) ¸ (z + 4 – i)

Again:

 f(-3i) = {(-3i +2 -3i)/ (-3i + 4 – i)} =  (2 – 6i)/ (4 – 4i) = 1 –  ½ i

2. Find f(2i -3) for f(z) = (z + 3)²(z – 5i)² 

è       f(2i -3) =  {(2i -3)+3}² (2i – 3 – 5i)² =  {(-4)(18i)} = -72i

3. Let f(z) = ln r + i(q)  where r = êz ê and q = Arg(z)

Find f(1):

f(1) = ln 1 + i Arg(1)  but we know that q  = p/4 for Arg (1)

Then:

f(1) = ln 1 + i(p/4)

4. Find:  f(i p/4)  for f(z) = exp(x) cos(y) + i(exp(x)sin(y)

Here, let z = r exp(i q)  then q = p/4

And exp(i p/4)=  cos(p/4) + isin(p/4) with r = 1

Therefore:

f (z) = exp(1)cos(p/4) +i(exp(1)sin(p/4) 

f(z) = exp[(cos(p/4) +i sin(p/4)] =  exp{Ö2/2 + iÖ2/2}

 f(z) =   1.922 + 1.922i

Functions of a complex variable can also be written in the form:

f(x + iy) = u + iv

and since u,v depend on x and y, they can be considered as real functions of the real variables x and y such that:

u = u(x,y) and v = v(x,y)

Example: Write f(z) = z²  in the form f(z) = u(x,y) + iv(x,y)

If z = (x + iy) then:   z² = (x + iy)² = x² + i2xy  - y²  

=  (x² – y²) + i2xy

The last step above shows how the complex function is separated into two parts, one with the factor i, the other without. The one with the factor applies to the function v(x,y) so:

v(x,y) = 2xy   

While:  :   u(x,y) = x² + y²

Conversely, of course, one can be given the functions u(x,y) and v(x,y) then be asked to find f(z), e.g. in terms of z and-or its complex conjugate, z*.

Example:

Given u(x,y) + iv(x,y) =  = 4x² + i4y²

 Find f(z,z*)

Again, let:  z = x + iy, and z* = x – iy

Adding:

z + z* = (x + iy) + (x -  iy) = 2x

So we see: x = (z + z*)/2

Now, subtracting: (z – z*) = [x + iy – x + iy] =  i2y

So: y = (z – z*)/ 2i

 

Since we have both x and y we can now formulate the function f(z,z*):

f(z,z*) = 4[(z + z*)/2]²  + i4[(z – z*)/ 2i]²

It’s important to note that a polar form is also used, viz.

f (z) = f(r exp(i(q)) = u(r, q) + iv(r, q)

Example: Express f(z)= z²   in polar form.

z² = r² exp(i2(q)) = r² (cos (2q) + isin(2 q))

Therefore:

u(r, q) =  r²(cos (2q)    and:

v (r, q) =    r²(sin (2q)) 

These manipulations also allow one to see how to solve complex function equations..

For example: Solve for x and y in the equation:

3i +  6x   =   2  -   iy

As we've seen with the examples, one can work with the real part and the complex part separately.

Then:   6x = 2   so   x = 2 /6   =  1/ 3

And:    3i   =    -iy   or:    y  =   3i / (-i)  =   -3

Another example:

Solve for x and y:

4x² + i4y²    = 2 x   + yi

Then:     4x² =   2x   So:   x  =  2/ 4  =  1/2

And:  

i4y²    =   yi

So:    y   =   i /   4i     =   1/ 4   

Problems for the Math Maven:

1)  (a) Given u(x,y) + iv(x,y) = 2x² + i2y²  

find f(z,z*)

(b)  Express f(z)= z² + z – 3 in polar form

2) Solve for x and y:   8x 2 + 3iy  -  4    = 8y   – 4iy

3) Solve: (z+1) 3  = z 3

Math Revisited: Complex Algebra

One graphical method of representing complex numbers. For example, here C = 4 + 3i


Back in 2010 and earlier this year we examined multiple features of complex analysis, including the nature of complex numbers, complex functions, simple manipulations -  for example,  getting a complex number into polar form, as well as complex roots (e.g. Ö2i), the Cauchy -Riemann equations and residue calculus.  A basic area includes complex algebra, or solving algebraic problems using the principles of complex numbers.  Below are a number of selected problems to challenge math -inclined readers looking for more than Instagram stuff or Trump's idiotic tweets:

1)   Let f(z) = ln r + i(q)  where r = êz ê and q = Arg(z)

Find f(1)

2)   Find f(2i -3) for f(z) = (z + 3)²(z – 5i)² 

3)   Find f(2i) for f(z) =  - 3z²


4)   Let f(z) = e ^(-3z)

Find the real and imaginary parts of the function f(z)

5)  Find all solutions for cos (z) =  5

6) Solve for z if:  sin z =  i sinh  1

7)  If   ln 1  = ln 1 +  ln 0i

Solve for z   

8)   Find: f(-3i) for f(z) = (z + 2 – 3i) ¸ (z + 4 – i)

9)  Find f(1+i) for: f(z) = 1/ (z² + 1)


10)   Solve: (z + 1)³ = z³

Math Revisited: Complex Analysis (1)

In advanced physics, including plasma physics and quantum mechanics, the importance of complex numbers and analysis can't be over emphasized. Consider the diagram shown with three complex numbers identified.   To extend the generality of complex numbers and enhance their applicability, it's useful to write them in what's called "polar form". 

A critical part is finding the angle shown, referred to as the argument. We can see from the diagram that the angle Θ may be found using:

arctan (y/x) = arctan(3/4) = 36.8 deg

Thus, Θ = 36.8 degrees is the argument


Now, any complex number (x + iy) may be written in polar form:

  x + iy = r(cos (Θ) + isin(Θ))


To get r:

r = [x² + y²]^1/2 =  [4² + 3²]^1/2 = [25]^1/2 = 5

Therefore we may write:  

 (x + iy) = 5(cos (36.8) + isin(36.8))

Note there is also the abbreviated function (based on the combo of sine and cosine):

cis (Θ) = cos (Θ) + isin(Θ)

So we can finally write:

C = r cis(Θ) = 5 cis (36.8)

We look now at the vectors A and B, which we’ll henceforth call z1 and z2 to be consistent with complex notation. Our eventual goal will be to find the resultant, which will come in the next installment. In the meantime we will be working toward showing the multiplication and division of two complex forms, call them z1 and z2: 

e.g.  [z1 + z2]

From the diagram:

A= z1 = -2 + 2i

B = z2 = -2 -3i

So: z1 = x1 + iy1

And  arg(z1) = arctan(y1/x1) = arctan (-2/2) = arctan(-1)

So (Θ1) = -45 degrees = -π /4

Now find r1:

r₁ =[x₁² + y₁²]^1/2 =  [1 + 1]^1/2 = Ö2   Therefore:


z1 = Ö2 (cos(-45) + isin(-45)) = Ö2 cis(-45)

We now turn to the vector B which is:  z2 = x2 + iy2= -2 -3i

then: arg(z2) = arctan(y2/x2) = arctan (-3/-2) = arctan (3/2) = 56.3 deg

While:

r₂ = [x₂² + y₂²]^1/2 =  [(-2)² + (-3)²]^1/2 = [13]^1/2 =  3.6


Therefore:

z2 = 3.6(cos(56.3) + isin(56.3) = 3.6 cis(56.3)

Now, how do we obtain the complex product: [z1•z2]?

We have that:

[z1•z2] = (z1•z2) cis(arg(z1) – arg(z2))

But:

(z1•z2) = Ö2 (3.6) = 5.1

And:

arg(z1) – arg(z2) = (-45) – (56.3) = -101.3

so that:


[z1•z2] = 5.1 cis(-101.3) = 5.1 (cos (-101.3) + isin(-101.3))


[z1•z2] = 5.1((-0.195) + i(-0,98))


[z1•z2] = 0.99 + 0.98i

To get the resultant: z1 + z2 = z3:

A + B = z1 + z2 =[ (-2 + 2i) + (-2 – 3i)] = -4 –i

In any case: x3 + iy3 = - 4 – i

Complex Division:

Let's say we want to divide:

z1 = Ö2(cos(-45) + isin(-45)) = Ö2 cis(-45)

by

z2 = 3.6(cos(56.3) + isin(56.3)) = 3.6 cis(56.3)

In all such cases of complex division we require that the z, r in the denominator not be zero.

Thus:

(z₁/z₂) =  (r₁ cis(q₁)/ r₂ cis(q₂)) = (r₁/ r₂) cis (q₁ –  q₂)

 

Now: (r1/ r2) = (1.414/ 3.6) = 0.39

And we saw previously:

(q₁ – q₂)   = arg(z₁) – arg(z₂) = (-45) – (56.3) = -101.3

Thus, the basic procedure for division entails dividing the lengths (r’s) and subtracting the angles    (q₁ – q₂).

So:
(z₁/ z₂) =  0.39 (cos (-101.3) + isin(-101.3))

= 0.39((-0.195) + i(-0.98)) = -0.07 + 0.38i
 

What about?   (1 + i)  ¸  Ö3  – i

The first order of business is to get dividend and divisor each into polar form, specifically as a (cis) function. 

Then (1 + i) = z₁  = x1 + iy1, so arg(z₁ ) = arctan (y1/x1)
Further:

arctan (y1/x1) = arctan (1/1) = arctan (1) so q₁ = 45 deg

What about r1?

r1= [1² + 1²]^1/2 =  Ö2 = 1.4

so z₁ = 1.4 [cos (45) + isin(45)] = 1.4 cis(45)
 

Now: z₂ = Ö3  – i

So arg(z₂) = arctan(y₂/x₂) =  arctan(-1/ Ö3) so q₂

= (-30 deg)

r₁/r₂ = Ö2/ 2 

Then subtract angles: [(q₁ – q₂) ] = {(45 deg) – (-30 deg)} = 75 degrees

So the end result of the division is:

(z₁/z₂)  =  Ö2/ 2   cis(75) = Ö2/ 2  {cos(75) + isin(75)}

= 0.707{cos(75) + isin(75)}

Since cos(75) = 0.258 and sin(75) =0.966, we have:

(z₁/z₂)   = 0.707[(0.258) + i(0.966)] = 0.183 + 0.683i

Another very convenient way to express complex numbers is in the exponential form.

Thus, we can write: cos(q) + isin(q) = r exp (iq )

Thus, the previous numbers we divided (z₁ and z₂) may be expressed:

z₁ = Ö2 [cos (45) + isin(45)] = Ö2 exp (i p/4)

z2 = 2[cos(-30) +isin(-30)] = 2 exp(i (-p/6))

Practice Problems:

1) Express each of the following end results in the form:   r exp(iq):

a) (2 + 3i)(1 – 2i)

b) (1 + i) (1- i)

c) (1 + Ö-3)²

2) Plot the results of (b) and (c) on the same Argand diagram and obtain the resultant. Check