AcademyComplex Equations

Academy

Linear Complex Equations

Level 1 - Math I (Physics) topic page in Complex Equations.

Principle

A linear complex equation is a first-degree equation for an unknown complex number. It has the same algebraic shape as a real linear equation, but the coefficients and the solution may be complex numbers.

Linear complex equation
\[az + b = 0,\quad a,b,z\in\mathbb C,\quad a\ne 0\]

Because every non-zero complex number has a reciprocal, a linear complex equation with \(a\ne0\) has exactly one complex solution.

Notation

  • \(z\) is the unknown complex number.
  • \(a\) is the coefficient of \(z\).
  • \(b\) is the constant term.
  • \(\overline{a}\) is the complex conjugate of \(a\).
  • \(|a|\) is the modulus of \(a\).

Method

Start with the equation:

Start
\[az + b = 0\]

Subtract \(b\) from both sides:

Isolate the z term
\[az = -b\]

Divide by \(a\), which is allowed because \(a\ne0\):

Linear solution
\[z = -\frac{b}{a}\]

If the denominator is complex, write the quotient in standard form by multiplying top and bottom by the conjugate of the denominator:

Conjugate division
\[\frac{w}{a}=\frac{w\overline{a}}{a\overline{a}}=\frac{w\overline{a}}{|a|^2}\]

The denominator \(|a|^2\) is real and positive, so the final expression can be separated into real and imaginary parts.

Rules

  • A degree-one complex equation has one solution when the coefficient of \(z\) is not zero.
  • If \(a=0\) and \(b\ne0\), the equation has no solution.
  • If \(a=0\) and \(b=0\), every complex number is a solution.
  • Complex numbers are not ordered, so solving does not involve choosing a larger or smaller solution.

Examples

Solve:

Example equation
\[(2-i)z+(1+3i)=0\]

First isolate \(z\):

Example isolate
\[(2-i)z=-(1+3i)=-1-3i\]

Divide by \(2-i\):

Example quotient
\[z=\frac{-1-3i}{2-i}\]

Multiply top and bottom by \(2+i\):

Example rationalise
\[z=\frac{(-1-3i)(2+i)}{(2-i)(2+i)}\]

Compute the numerator and denominator:

Example expansion
\[(-1-3i)(2+i)=-2-i-6i-3i^2=1-7i\]
Example denominator
\[(2-i)(2+i)=4-i^2=5\]

Therefore:

Example answer
\[z=\frac{1-7i}{5}=\frac15-\frac75 i\]

Checks

  • Check that the coefficient of \(z\) is not zero before dividing.
  • Use the conjugate of the denominator, not the numerator, when simplifying a complex quotient.
  • Substitute the result back into the original equation to catch sign errors.
  • Keep the final answer in standard form \(x+iy\) unless polar form is requested.