In one line
Complex numbers follow ordinary algebra with the single extra rule \(i^2=-1\); the conjugate \(\bar z=x-iy\) is the tool that removes \(i\) from a denominator.
Explained simply
Treat \(i\) like a variable such as \(a\) in school algebra, expand normally, and wherever you see \(i^2\) replace it by \(-1\). Division is like rationalising \(\dfrac{1}{\sqrt2+1}\): you multiply top and bottom by a partner that makes the bottom real. For complex numbers that partner is the conjugate.
Formally, for \(z_1=a+ib\), \(z_2=c+id\): \(z_1\pm z_2=(a\pm c)+i(b\pm d)\), \(z_1z_2=(ac-bd)+i(ad+bc)\). The conjugate of \(z=x+iy\) is \(\bar z=x-iy\) (reflection in the real axis), and \(z\bar z=x^2+y^2=|z|^2\), a real number.
How it works
- Addition/subtraction: combine real parts and imaginary parts separately.
- Multiplication: expand like binomials, replace \(i^2=-1\), group real and imaginary terms.
- Division \(\dfrac{z_1}{z_2}\): multiply numerator and denominator by \(\bar z_2\); the denominator becomes \(|z_2|^2\).
- Powers of \(i\) repeat in a cycle of four: \(i,\,-1,\,-i,\,1\). Divide the exponent by 4 and use the remainder.
Key formula
- \(|z|=\sqrt{x^2+y^2}\), \(z\bar z=|z|^2\)
- \(\overline{z_1z_2}=\bar z_1\bar z_2\), \(\overline{z_1+z_2}=\bar z_1+\bar z_2\)
- \(|z_1z_2|=|z_1||z_2|\), \(\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}\)
- \(\arg(z_1z_2)=\arg z_1+\arg z_2\), \(\arg\left(\dfrac{z_1}{z_2}\right)=\arg z_1-\arg z_2\) (up to adding multiples of \(2\pi\))
- \(\mathrm{Re}\,z=\dfrac{z+\bar z}{2}\), \(\mathrm{Im}\,z=\dfrac{z-\bar z}{2i}\)
All quantities here are pure numbers; there are no units.
Quick example
Simplify \(\dfrac{3+2i}{1-i}\). Multiply top and bottom by \(1+i\):
\(\dfrac{(3+2i)(1+i)}{(1-i)(1+i)}=\dfrac{3+3i+2i+2i^2}{1+1}=\dfrac{1+5i}{2}=\dfrac12+\dfrac52 i\).
Check with moduli: \(|3+2i|/|1-i|=\sqrt{13}/\sqrt2=\sqrt{6.5}\), and \(|\tfrac12+\tfrac52i|=\sqrt{0.25+6.25}=\sqrt{6.5}\). They agree.
Common mistakes
- Writing \(|z|^2=x^2-y^2\); it is always \(x^2+y^2\).
- Multiplying only the denominator by the conjugate and forgetting the numerator.
- Assuming \(|z_1+z_2|=|z_1|+|z_2|\); only the inequality \(|z_1+z_2|\le|z_1|+|z_2|\) is true.
How MU asks it
This is the self-learning basics topic, so it is not usually a separate question. It appears inside other problems: "If \(\dfrac{1}{\alpha+i\beta}+\dfrac{1}{a+ib}=1\)..., find..." or as the simplification step before applying De Moivre's theorem. For full marks, show the conjugate multiplication line and write the answer clearly as real part + \(i\) imaginary part.