Calculate $(4-3i)-(1+2i)$.
- Distribute the subtraction: $4-3i-1-2i$.
- Combine to obtain $3-5i$.
Why does subtraction affect both components?
The minus sign acts on both components of the second number.
Complex numbers in Cartesian form
Read complex numbers as coordinates, then use conjugates to multiply and divide.
Understand the idea
A complex number has two components.
In $z=a+bi$, $a$ gives the horizontal position on the real axis; $b$ gives the vertical position on the imaginary axis. A point or a vector from the origin represents the same complex number.
$z=2+3i\quad\longleftrightarrow\quad\text{point }(2,3)$
See it visually
Imagine reflecting the point in the real axis. Predict which component changes before you try it.
Interactive guide
Check the logic
What will conjugation do?
Explore the result
Reflection keeps the real component and reverses the imaginary component.
Change a, then b. Compare z with its conjugate: which component always stays the same?
At a = 2 and b = 3, z is (2,3) and its conjugate is (2,−3). The real component stays 2; the imaginary component changes from 3 to −3.
Core notation
Cartesian form
$z=a+bi,\quad a,b\in\mathbb R$
Real and imaginary parts
$\operatorname{Re}(z)=a,\quad\operatorname{Im}(z)=b$
Conjugate
$\overline{a+bi}=a-bi$
Predict an addition
Check the logic
Starting from 2+3i, predict the movement when adding 2−2i.
Explore the result
Add 2 to the horizontal coordinate and subtract 2 from the vertical coordinate: (2,3) → (4,1), so the result is 4+i.
Compare your two directions with the movement shown.
Operations
Addition combines matching components. Multiplication is ordinary expansion, with one extra rule: $i^2=-1$.
Group the real terms: $ac+bdi^2=ac-bd$. Group the imaginary terms: $adi+bci=(ad+bc)i$.
Addition
$(a+bi)+(c+di)=(a+c)+(b+d)i$
Multiplication: expand
$(a+bi)(c+di)=(ac-bd)+(ad+bc)i$
Division: make the denominator real
The goal is to remove $i$ from the denominator. Choose its conjugate because $(c+di)(c-di)=c^2+d^2$. Multiplying the numerator by the same factor preserves the quotient.
Division: use the conjugate
$\dfrac{a+bi}{c+di}=\dfrac{(a+bi)(c-di)}{c^2+d^2}$
For $c+di\ne0$, $c^2+d^2>0$.
For $c+di\ne0$, $c^2+d^2>0$.
Checkpoint
Correct. The imaginary part is the real coefficient of i.
Not quite. That is the imaginary term, not the coefficient called Im(z). That is the real part.
Show explanation
The imaginary part is the real coefficient of i.
Worked examples
Example 1
Combine matching components
Example 2
Use the square of i
Expand $(2+i)(3-2i)$.
Check the logic
After expansion: $6-4i+3i-2i^2$. What should happen to $-2i^2$?
Show the working
Because i² = −1, −2i² = −2(−1) = 2.
Now compare that real term with the completed working.
- $6-4i+3i-2i^2$.
- Use $i^2=-1$: $6-i+2=8-i$.
Why does i² create a real term?
Because $i^2=-1$, multiplying two imaginary factors produces a real number: $(bi)(di)=bd i^2=-bd$. Here $i(-2i)=-2i^2=2$.
Example 3
Make the denominator real
Write $(3+i)/(1-i)$ in Cartesian form.
- Multiply top and bottom by $1+i$.
- The numerator is $3+4i+i^2=2+4i$; the denominator is $1-i^2=2$.
- The quotient is $1+2i$.
Why does the conjugate make the denominator real?
Multiplying both parts of a fraction by the same nonzero number preserves its value; the conjugate makes the denominator real.
In the exam
Quick check
Your score is not saved.
The defining rule is i² = −1.
Show explanation
The defining rule is i² = −1.
Add real parts and add the coefficients of i.
Show explanation
Add real parts and add the coefficients of i.
Only the imaginary component changes sign.
Show explanation
Only the imaginary component changes sign.
1 − i² = 2.
Show explanation
1 − i² = 2.
Multiply both sides by z. The conjugate product must be 1, so the reciprocal has no extra real scale factor.
Show explanation
Multiply both sides by z. The conjugate product must be 1, so the reciprocal has no extra real scale factor.
Lesson Rewards
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