Calculate with complex numbers in Cartesian form

Use quadratic roots and conjugate factors

First solve real-coefficient quadratics with negative discriminant and form the real quadratic factor forced by a known nonreal root. These calculations support division, residual equations and multiplicity analysis in the full polynomial

Use the discriminant over the stated domain

For real A, B and C with a nonzero leading coefficient, define the discriminant

Two distinct real roots

One real value with multiplicity two

Two distinct non-real conjugates

Use the quadratic formula in each case and state the solution domain

Find the discriminant D of the quadratic

Extract the imaginary factor carefully

For a negative real discriminant, extract the imaginary factor before simplifying the positive real square root. Do not apply the product rule for square roots indiscriminately to negative real inputs

1. 1Quadratic formula The coefficients are real and the leading coefficient is nonzero \(\displaystyle \begin{gathered}D=B^2-4AC<0\\z=\frac{-B\pm i\sqrt{-D}}{2A}\end{gathered}\)
2. 2Example Substitute the coefficients of the displayed quadratic \(\displaystyle \begin{gathered}2z^2-4z+5=0\\D=16-40=-24\\z=\frac{4\pm i\sqrt{24}}4\end{gathered}\)
3. 3Simplify Extract the positive square factor \(\displaystyle \begin{gathered}\sqrt{24}=2\sqrt6\\z=1\pm\frac{\sqrt6}{2}i\end{gathered}\)

Equation

\(\displaystyle 2z^2-4z+5=0\)

Root form

\(\displaystyle h\pm k\sqrt d\,i,\quad k\in\mathbb Q_{>0}\)

Radicand

\(\displaystyle \begin{gathered}d>0,\quad d\in\mathbb Z\\d\text{ is square-free}\end{gathered}\)

Enter the common real part in the stated root form

Equation

\(\displaystyle 2z^2-4z+5=0\)

Root form

\(\displaystyle h\pm k\sqrt d\,i,\quad k\in\mathbb Q_{>0}\)

Radicand

\(\displaystyle \begin{gathered}d>0,\quad d\in\mathbb Z\\d\text{ is square-free}\end{gathered}\)

Enter the positive coefficient in the stated root form

For z²+9=0 over C, which set is complete?

A conjugate pair gives a real quadratic

A real-coefficient polynomial with a known nonreal root also has its conjugate. Their monic quadratic is a factor, but the full polynomial and multiplicities need further information

1. 1General pair Multiply the two linear factors \(\displaystyle \begin{gathered}a,b\in\mathbb R,\quad b\ne0\\(z-a-bi)(z-a+bi)\\=(z-a)^2-(bi)^2\\=(z-a)^2+b^2\end{gathered}\)
2. 2Example Use the given conjugate pair \(\displaystyle \begin{gathered}2\pm3i\\(z-2-3i)(z-2+3i)\\=(z-2)^2+9\\=z^2-4z+13\end{gathered}\)