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Edexcel A Level Further Maths: Core Pure:復習筆記1.1.2 Solving Equations with Complex Roots

Category: A-level課程, 國際課程, 教材筆記, 福利干貨 Date: 2022年10月17日上午10:29

Complex numbers?provide solutions for quadratic equations which have?no real roots

8-1-4-complex-roots-of-polynomials-diagram-1

Complex roots occur when solving a quadratic with a?negative discriminant
- This leads to square rooting a negative number

We solve an equation with complex roots in the same way we solve any other quadratic equations
- If in the form ?we can?rearrange?to solve
- If in the form ?we can?complete the square?or use the?quadratic formula
We use the property ?along with a manipulation of surds

When?the?coefficients?of the quadratic equation are?real, complex roots occur in?complex conjugate?pairs
- ?is a root of a quadratic with real coefficients then ?is also a root
When the coefficients of the quadratic equation are?non-real, the solutions will?not?be complex conjugates
- To solve these use the quadratic formula

Once you have your final answers you can check your roots are correct by substituting your solutions back into the original equation.
You should get 0 if correct! [Note: 0 is equivalent to ]

a) Solve the quadratic equation?z2?- 2z?+ 5 = 0 and hence, factorise?z2?- 2z?+ 5.

1-1-2-al-further-maths-complex-roots-of-polynomials

b) Given that one root of a quadratic equation is?z?= 2 – 3i, find the quadratic equation in the form?az2?+?bz?+?c?= 0, where?a, b,?and?c?∈ ?,?a?≠ 0.

1-9-3-ib-aa-hl-complex-roots-we-solution-1-b

We know that every?quadratic?equation has?two roots?(not necessarily distinct or real)
This is a particular case of a more general rule:
- Every polynomial equation, with real coefficients, of degree?n?has?n?roots
- The?n?roots are not necessarily all?distinct?and therefore we need to count any?repeated?roots that may occur individually
From the above rule we can state the following:
- A cubic equation of the form ?can have either:
  - 3?real?roots
  - Or 1?real?root and a complex?conjugate?pair

A quartic equation of the form ?will have one of the following cases for roots:
- 4?real?roots
- 2?real?and 2?nonreal?(a complex conjugate pair)
- 4?nonreal?(two complex conjugate pairs)
When a real polynomial of any degree has one complex root it will always also have the complex conjugate as a root

From there either
- Expand and compare coefficients to find
- Or use polynomial division to find the factor
Finally, write your?three roots?clearly

When asked to find the roots of any polynomial when we are given one, we use almost the same method as for a cubic equation
- State the initial root and its conjugate and write their factors as a quadratic factor (as above) we will have two unknown roots to find, write these as factors (z?- α) and (z?- β)
- The unknown factors also form a quadratic factor (z?- α)(z?- β)
- Then continue with the steps from above, either?comparing coefficients?or using?polynomial division
  - If using polynomial division, then solve the quadratic factor you get to find the roots α and β

If we are given a root of a polynomial of any degree in the form?z??=?p?+?qi
- We know that the complex conjugate,?z*?=?p?–?qi is another root
- We can write (z?– (p?+?qi)) and (?z?– (p?-?qi)) as two linear factors
  - Or rearrange into one quadratic factor
- This can be multiplied out with another factor to find further factors of the polynomial
For higher order polynomials more than one root may be given
- If the further given root is complex then its complex conjugate will also be a root
- This will allow you to find further factors

As with solving quadratic equations, we can substitute our solutions back into the original equation to check we get 0
You can speed up multiplying two complex conjugate factors together by
- rewrite (z?– (p?+?qi))(z?– (p?-?qi)) as ((z?–?p) -?qi))(?(z?–?p) +?qi))
- Then?((z?–?p) -?qi))(?(z?–?p) +?qi)) = (z - p)2?- (qi)2?=?(z - p)2?+ q2