IB MATHEMATICS Topic ; Complex Numbers – Solving Complex Simultaneous Equations

Complex simultaneous equations are in fact just normal simultaneous equations and can be solved in the usual way, by equating coefficients of one variable and eliminating, or by substitution.
To solve the complex simultaneous equations
(1)
(2)
Rearrange (1) to give(3) and substitute this into (2) to give

Partially expanding brackets gives


Factorising with w and taking the first term on the left over to the right hand side gives

Hence
Substituting this into (3) gives

IB MATHEMATICS Topic : Complex Numbers – Solving Absolute Value Complex Equations

Absolute value equations typically do not have single solutions, or even a set of solutions which can be listed. Typically, the solution describes a curve in the complex plane. To take a very simple example, the equationhas the solution given in polar form asor in cartesian form aswith
Often it is easiest to find the solution in cartesian form by substituting z=x+iy and collecting real and imaginary terms, squaring and adding them to give a real number.
Example: Solve
Write the equation asand multiply byto give(1)
Now substitute z=x+iy.


Substitute these two expressions into (1) to obtain
Square both sides to give
Now multiply out the brackets and collect like terms.


Divide by 3 and complete the square.



This is the equation of a circle with centreand radius

IB MATHEMATICS Topic : Complex Numbers – de Moivre's Theorem

De Moivre's theorem states that forwhere

The theorem is easy to prove using the relationship Raising both sides of this expression to the power ofgives

The theorem is useful when deriving relationships between trigonometric functions. For example, we can obtain polynomial expressions for sin n %theta and cos n %theta for any n using de Moivre's theorem.

Example: Derive expressions forandusing de Moivre's theorem.

(1)

Expanding the left hand side using the binomial theorem gives

(2)

Equating real coefficients of (1) and (2) gives respectively

Useto give Simplifying this expression gives

Equating imaginary coefficients of (1) and (2) gives respectively

Useto give Simplifying this expression gives

(1)

Expanding the left hand side using the binomial theorem gives

The real and imaginary parts of this expression are

Equating real parts gives

Useto give

This simplifies to

Equating imaginary parts gives

Useto give

This simplifies to