3FGSolvingCmplxEqns


 * Solving Complex Equations **

Equations in C can be solved in the same way as equations in R.

To solve,
 * you can factorise and use the Null Factor Law (See Example 1 below)
 * To factorise it may be necessary to complete the square and introduce **i** 2 = –1
 * __or__**
 * use the Quadratic Formula and introduce **i**. (see Example 2 below)


 * Example 1 **

math \text{Factorise } \; z^3-4z^2+9z-36 \\. \\ \text{and hence solve } \; z^3-4z^2+9z-36=0 math


 * Solution:**

... ... P(4) = 0 ... ... so (z – 4) is a factor


 * Use long division or a shortcut or your calculator to factorise}

math \\ . \qquad z^3-4z^2+9z-36=0 \\. \\ . \qquad \left( z^2+9 \right) \left( z-4 \right) = 0 math


 * If factorising in the real field, this would be fully factorised,
 * but in the complex field, we keep going by introducting i 2 = –1
 * then we can use the difference of 2 squares rule

math \\ . \qquad \left( z^2-9 \, i^2 \right) \left( z-4 \right) = 0 \\. \\ . \qquad (z+3 \, i)(z-3 \, i)(z-4) = 0 math


 * This is now __factorised__ in C.
 * From this, we see that the solutions to the __equation__ are:

math \\ . \qquad z^3-4z^2+9z-36=0 \\. \\ . \qquad (z+3 \, i)(z-3 \, i)(z-4) = 0 \\. math math . \qquad z = -3i, \; \; z = +3i, \; \; z = 4 math

More examples

Factorising and Solving on the calculator

You can find the factors of a polynomial over C on the classpad by setting the calculator to **CPLX** mode Select **Action** menu, **Transformation** submenu, __**factor**__ Then insert your desired function

math . \qquad \textbf{factor } (z^3-4z^2+9z-36) math

You can solve equations over C by setting the calculator to **CPLX** mode and using the __**solve**__ command. __**solve**__ is in the **ACTION** menu, **ADVANCED** submenu and in the **EQUATION** submenu.

math . \qquad \textbf{solve } (z^3-4z^2+9z-36 =0, \; z) math

Solving Using the Quadratic Formula


 * Example 2 **

math \text{Solve } \; 2z^2-2z+3=0 math


 * Solution:**
 * Use Quadratic Formula
 * with a = 2, b = –2, c = 3

math \\ . \qquad z=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\. \\ . \qquad z=\dfrac{2 \pm \sqrt{(-2)^2 - 4 \times 2 \times 3}}{2 \times 2} \\. \\ . \qquad z=\dfrac {2 \pm \sqrt{-20}}{4} math

Now simplify the surd, introduce i and separate fraction into real and imaginary parts

math \\ . \qquad z=\dfrac {2 \pm \sqrt{-1} \times \sqrt{4} \times \sqrt{5}}{4} \\. \\ . \qquad z=\dfrac {2}{4} \pm \dfrac {2 \sqrt{5} \, i}{4} \\. \\ . \qquad z=\dfrac {1}{2} \pm \dfrac {\sqrt{5} \, i}{2} math

The Fundamental Theorem of Algebra

Any polynomial equation P(z)=0 has a solution when solved over C,

therefore:

For any polynomial P(z) of degree **n**
 * P(z) will have **n** linear __factors__ over C (some may be repeated)
 * P(z)=0 will have **n** linear __solutions__ over C


 * a quadratic always has 2 factors,
 * a cubic always has 3 factors
 * etc

If at least one __coefficient__ of P(z) is complex, then at least one solution of P(z)=0 will be a complex number.

Note
 * Sometimes the solutions to P(z) = 0 are called __zeros__
 * because they are values of z where P(z) is zero.
 * Solutions to __any__ equation are called __roots__
 * the reason for that is unclear.

For another site that explains this idea, go here: MathsIsFun

The Conjugate Root Theorem

If P(z) is a polynomial with only __real__ coefficients and ... ... z = a + bi .. is a solution to P(z) = 0

then the conjugate is also a solution

... ... z = a – bi is also a solution.


 * any complex solutions come in conjugate pairs, provided all the coefficients are real


 * Example 3 **

Given that **//z = –3 + 2i//** is a solution to: math . \qquad z^3+5z^2+7z-13=0 math

find the other solutions.


 * Solution:**


 * By the __Fundamental Theorem of Algebra__,
 * there will be **3** solutions.


 * By the __Conjugate Root Theorem__,
 * since the coefficients are all real,
 * another solution will be //**z = –3** – **2i**// ... {the complex conjugate of the first solution}


 * The third solution __must__ be real
 * since it cant be matched with a conjugate pair
 * P(1) = 0,
 * so the third solution is **//z = 1//**
 * The factor is **//(z – 1)//**

So the solutions are: **z = 1, –3 ± 2i**


 * Don’t forget to reverse the sign when forming the factors **!

OR (alternate method to find 3rd solution)

The real solution can also be found by multiplying the known factors (not the roots) and then dividing into P(z).

math \\ . \qquad (z+3-2 \, i)(z+3+2 \, i) \\. \\ . \qquad = z^2 + 6z + 13 math

Long Division: math z^2 + 6z + 13 \overset{\displaystyle {z-1} \qquad \qquad \; \, .} {\overline { \left) z^3+5z^2+7z-13 \right. }} \\ . math math \\ . \qquad \qquad \quad \; \underline {z^3 + 6z^2 + 13z} \\ . \\ . \qquad \qquad \qquad \, -z^2 - 6z - 13 \\ . \\ . \qquad \qquad \qquad \; \, \underline {-z^2 - 6z - 13} \\ . \\ . \qquad \qquad \qquad \qquad \qquad \qquad 0 math

hence the real factor is **z – 1**

More Examples

Solving Equations on the Classpad

You can find real solutions by graphing the equation on your calculator and locating the x-intercepts.

Or, you can find __all__ solutions by using your calculator to solve the equation in CPLX mode.

Select **Action** menu, **Advanced** submenu, **solve** Type in the equation, then a comma, then the variable (**z**) and finally close the brackets.

For the example above, your entry should look like this: math . \qquad \textbf{solve } \big( z^3 + 5z^2 + 7z - 13 = 0, \; z \big) math

Finding Roots

math \\ \text{To solve } z^n = a \\. \\ \text{we have to find the nth root } z = \sqrt[n]{a} math

This can be done using De Moivre's Theorem

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