Edexcel A Level Further Maths: Core Pure:復(fù)習(xí)筆記1.1.2 Solving Equations with Complex Roots
Solving Quadratic Equations with Complex Roots
What are complex roots??
- Complex numbers?provide solutions for quadratic equations which have?no real roots
- Complex roots occur when solving a quadratic with a?negative discriminant
- This leads to square rooting a negative number
How do we solve a quadratic equation with complex roots?
- 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
- If in the form
- 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
How do we find a quadratic equation given a complex root?
- We can find the equation of the form
?if you are given a complex root in the form 
- We know that the complex conjugate
?is another root, - This means that
?are?factors?of the quadratic equation - Therefore
- Writing this as
?will speed up expanding
- Writing this as
- Expanding and simplifying gives us a quadratic equation where b and c?are?real?numbers
- We know that the complex conjugate
Exam Tip
- 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
]
Worked Example
a) Solve the quadratic equation?z2?- 2z?+ 5 = 0 and hence, factorise?z2?- 2z?+ 5.
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.
Solving Polynomial Equations with Complex Roots
How many roots should a polynomial have?
- 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 cubic equation of the form
- 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
How do we solve a cubic equation with complex roots?
- Steps to solve a cubic equation with complex roots
- If we are told that
?is a root, then we know 
?is also a root - This means that
?and 
?are?factors?of the cubic equation - Multiply the above factors together gives us a quadratic factor of the form
- We need to find the third factor
- Multiply?the factors and?equate?to our original equation to get
- If we are told that
- From there either
- Expand and compare coefficients to find
- Or use polynomial division to find the factor
- Expand and compare coefficients to find
- Finally, write your?three roots?clearly
How do we solve a polynomial of any degree with complex roots?
- 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
How do we solve polynomial equations with unknown coefficients?
- Steps to find unknown variables in a given equation when given a root:
- Substitute?the given root?p?+ qi into the equation f(z) = 0
- Expand?and?group?together the?real?and?imaginary?parts (these expressions will contain our unknown values)
- Solve?as simultaneous equations to find the unknowns
- Substitute?the values into the?original?equation
- From here continue using the previously described methods for finding other roots for the polynomial
How do we factorise a polynomial when given a complex root?
- 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
Exam Tip
- 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
Worked Example
Given that one root of a polynomial p(x) =?z3?+?z2?–?7z?+?65?is 2 – 3i, find the other roots.
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