Operations with Complex Numbers

Add and subtract corresponding components

Add real parts to real parts and imaginary coefficients to imaginary coefficients:

For subtraction, distribute the minus sign across the entire second number:

The change from $+5i$ to $-5i$ is as important as the change from $-1$ to $+1$. Component notation can help organise the calculation, but it does not mean every complex operation is performed component by component.

Multiply by distributing first

Expand all four products before simplifying:

Since $i^2=-1$, the last term becomes $+3$, giving

In general,

Multiplying coordinate components to obtain $(ac,b d)$ would omit the cross terms and use the wrong real contribution. Complex multiplication is not the componentwise multiplication of two ordered pairs. The Complex Plane interprets it as rotation and scaling instead.

Use a conjugate to make the denominator real

To divide by $c+di\ne0$, multiply numerator and denominator by $c-di$. Their denominator product is $c^2+d^2$, which is positive when $c,d$ are real and not both zero.

For example,

The numerator is $3+6i+4i+8i^2=-5+10i$, and the denominator is $1+4=5$. Therefore

Check by multiplying back: $(-1+2i)(1-2i)=3+4i$. This validates the whole quotient rather than only checking a copied denominator.

The general expression is

provided $c+di\ne0$. The formula follows from the same expansion; it is not necessary to memorise it instead of understanding the conjugate method. Compare the standard complex arithmetic formulas.

Reduce powers with a genuine cycle

The powers of $i$ repeat in a four-step cycle. Since $2026=4\times506+2$,

Negative integer powers use reciprocals. Because $i(-i)=1$, we have $i^{-1}=-i$. For an expression containing more than one term, do not apply the cycle to the whole expression as if it were just $i$.

For example,

so $(1+i)^4=(2i)^2=-4$ and $(1+i)^8=16$. Repeated squaring can be shorter than multiplying out every factor.

Solve a complex linear equation

Solve $(2-i)z=5+i$. Since $2-i\ne0$, divide and multiply by its conjugate:

Thus $z=9/5+(7/5)i$. Multiplication by $2-i$ gives

as required. The equation has a unique solution because the coefficient is non-zero, just as for a real linear equation.

Find both square roots of a complex value

To solve $z^2=3+4i$, write $z=a+bi$ with real $a,b$. Equating components gives

The identity $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$ gives $(a^2+b^2)^2=25$. Since $a^2+b^2\ge0$, it equals 5. Adding and subtracting the equations yields $a^2=4$ and $b^2=1$.

The condition $ab=2$ requires matching signs. The solutions are therefore $2+i$ and $-2-i$, not all four possible sign pairings. Squaring verifies both. The factorization $z^2-(3+4i)=(z-(2+i))(z+(2+i))$ also shows that there are no other roots.

Keep conjugate and modulus identities distinct

For complex numbers, $z\overline z=|z|^2$, not generally $z^2$. Conjugation respects sums and products: $\overline{z+w}=\overline z+\overline w$ and $\overline{zw}=\overline z\,\overline w$. Modulus satisfies $|zw|=|z||w|$.

A pair of conjugate roots often produces a real quadratic factor. For $-2+3i$ and $-2-3i$,

For a polynomial with real coefficients, non-real roots occur with their conjugates. The reason is that conjugating the equation preserves every real coefficient and turns each occurrence of $z$ into $\overline z$.

Related topics

The Complex Plane turns addition, conjugation and multiplication by $i$ into visible geometric operations. Complex Numbers supplies the definitions and root conventions behind these calculations.