Argand Diagram Quiz — Name the Complex Number

Select the correct complex number \(a+bi\) for the plotted point.

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Complex Numbers — Key Notes & Practice

All complex numbers are written in the form a + bi, where i² = −1. Answers should be simplified to this form.

Key skills

• Add, subtract, multiply, and divide complex numbers.
• Use conjugates; know that \(z\overline{z}\) is real.
• Find the modulus as distance from the origin.
• Solve quadratics with complex roots.
• Solve equations by equating real and imaginary parts.

Jump to a section

• 1. Addition & subtraction
• 2. Multiplication
• 3. Conjugates
• 4. Division
• 5. Modulus
• 6. Quadratic equations
• 7. Solving equations by equating real & imaginary

1

Addition and subtraction

Combine real parts together and imaginary parts together.

Example: Simplify \( (3+2i) + (5-7i) \). Show solution

Real: \(3+5=8\)

Imaginary: \(2i-7i=-5i\)

Answer: \(8-5i\)

Practice: Simplify \( (4-3i) - (1+5i) \). Show solution

Real: \(4-1=3\)

Imaginary: \(-3i-5i=-8i\)

Answer: \(3-8i\)

2

Multiplication

Use an area/array model (2×2 grid), then combine real and imaginary parts (remember \(i^2=-1\)).

Example: \( (2+3i)(4-i) \) Show solution

Area model:

	\(4\)	\(-\,i\)
\(2\)	\(8\)	\(-2i\)
\(3i\)	\(12i\)	\(-3i^2\)

Real: \(8+(-3i^2)=8+3=11\)

Imaginary: \(-2i+12i=10i\)

Answer: \(11+10i\)

Practice: \( (1-2i)(3+i) \) Show solution

Area model:

	\(3\)	\(i\)
\(1\)	\(3\)	\(i\)
\(-2i\)	\(-6i\)	\(-2i^2\)

Real: \(3+(-2i^2)=3+2=5\)

Imaginary: \(i-6i=-5i\)

Answer: \(5-5i\)

3

Conjugates

The conjugate of \(a+bi\) is \(a-bi\). Then \(z\overline{z}=(a+bi)(a-bi)=a^2+b^2\) (real).

Example: For \(z=4-3i\), show \(z\overline{z}\) is real. Show solution

\((4-3i)(4+3i)=4^2-(3i)^2=16-9i^2\)

\(=16+9=25\in\mathbb{R}\)

Practice: If \(z=2+5i\), find \(z\overline{z}\). Show solution

\((2+5i)(2-5i)=2^2-(5i)^2=4-25i^2=29\)

4

Division

The conjugate of \(c+di\) is \(c-di\). Multiply numerator and denominator by \(c-di\):

\[ \frac{a+bi}{\,c+di\,} \;=\; \frac{(a+bi)(c-di)}{(c+di)(c-di)}. \]

Example: \(\displaystyle \frac{5+2i}{\,2-i\,}\) Show solution

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

\[ \frac{5+2i}{2-i}\cdot\frac{2+i}{2+i} =\frac{(5+2i)(2+i)}{(2-i)(2+i)}. \]

Numerator area model \((5+2i)(2+i)\)

	\(2\)	\(i\)
\(5\)	\(10\)	\(5i\)
\(2i\)	\(4i\)	\(2i^2\)

Denominator area model \((2-i)(2+i)\)

	\(2\)	\(i\)
\(2\)	\(4\)	\(2i\)
\(-i\)	\(-2i\)	\(-i^2\)

Numerator: \(10+5i+4i+2i^2=8+9i\)

Denominator: \(4+(-i^2)=4+1=5\)

Answer: \(\displaystyle \frac{8+9i}{5}= \frac{8}{5}+\frac{9}{5}i\)

Practice: \(\displaystyle \frac{3-4i}{\,1+2i\,}\) Show solution

Multiply top and bottom by the conjugate \(1-2i\):

\[ \frac{3-4i}{1+2i}\cdot\frac{1-2i}{1-2i} =\frac{(3-4i)(1-2i)}{(1+2i)(1-2i)}. \]

Numerator area model \((3-4i)(1-2i)\)

	\(1\)	\(-2i\)
\(3\)	\(3\)	\(-6i\)
\(-4i\)	\(-4i\)	\(8i^2\)

Denominator area model \((1+2i)(1-2i)\)

	\(1\)	\(-2i\)
\(1\)	\(1\)	\(-2i\)
\(2i\)	\(2i\)	\(-4i^2\)

Numerator: \(3-6i-4i+8i^2=-5-10i\)

Denominator: \(1-4i^2=1+4=5\)

Answer: \(-1-2i\)

5

Modulus

For \(z=a+bi\), the modulus is \(|z|=\sqrt{a^2+b^2}\) — the distance from the origin.

Example: Find \(|6-8i|\). Show solution

\(|6-8i|=\sqrt{6^2+(-8)^2}\)

\(\phantom{|6-8i|}=\sqrt{36+64}\)

\(\phantom{|6-8i|}=\sqrt{100}=10\)

Practice: Find \(|3+4i|\). Show solution

\(|3+4i|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

6

Quadratic equations with complex roots

When the discriminant \(b^2-4ac<0\), roots are complex.

Example: Solve \(x^2+2x+5=0\). Show solution

Quadratic formula: \(x=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}\)

Substitute: \(a=1,\;b=2,\;c=5\)

\(x=\dfrac{-(2)\pm\sqrt{\,(2)^2-4(1)(5)\,}}{2(1)}\)

\(=\dfrac{-2\pm\sqrt{-16}}{2}=\dfrac{-2\pm 4i}{2}=-1\pm2i\)

Practice: Solve \(x^2+4x+13=0\). Show solution

Quadratic formula: \(x=\dfrac{-b\pm\sqrt{\,b^2-4ac\,}}{2a}\)

Substitute: \(a=1,\;b=4,\;c=13\)

\(x=\dfrac{-(4)\pm\sqrt{\,(4)^2-4(1)(13),}}{2(1)}\)

\(=\dfrac{-4\pm\sqrt{-36}}{2}=\dfrac{-4\pm 6i}{2}=-2\pm3i\)

7

Solving complex equations by equating real and imaginary parts

Let \(z=a+bi\). Substitute and compare the real and imaginary parts separately.

Example: If \(2z+(3-i)=7+4i\), find \(a\) and \(b\). Show solution

\(2(a+bi)+(3-i)=(2a+3)+(2b-1)i\)

Compare with \(7+4i\)

Real: \(2a+3=7\Rightarrow a=2\)

Imaginary: \(2b-1=4\Rightarrow b=2.5\)

Therefore: \(z=2+2.5i\)

Practice: If \(z-(1+2i)=3-i\), find \(a,b\) where \(z=a+bi\). Show solution

\((a-1)+(b-2)i=3-i\)

Real: \(a-1=3\Rightarrow a=4\)

Imaginary: \(b-2=-1\Rightarrow b=1\)

© eMaths.ie — Complex Numbers Revision Notes

Argand Quiz — Which Has the Greater Modulus?

Two complex numbers \(z\) and \(w\) are shown. Each sits on a grey dotted circle centred at the origin. Choose the one with the larger modulus (radius).

z w dotted circle shows modulus

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Video Tutorials

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Addition and Multiplication

Division 

Argand Diagram and Modulus

Quadratic Equations with Complex Roots (Solutions)

Solving Complex Equations by Equations

Complex Numbers — Extra Questions

1) Addition & subtraction

Q. Simplify \( (7-5i) + (-3+2i) - (1-6i) \) to the form \( a+bi \).

Show solution

\[ \text{Real part: } 7+(-3)-1=3 \] \[ \text{Imaginary part: } (-5i)+2i+6i=3i \] \[ \boxed{\,3+3i\,} \]

2) Multiplication — area model

Q. Find \( (3+2i)(4-i) \).

Show solution

Area model:

	\(4\)	\(-\,i\)
\(3\)	\(12\)	\(-\,3i\)
\(2i\)	\(8i\)	\(-\,2i^{2}\)

\[ \text{Real terms: } 12 + (-2i^{2}) = 12 + 2 = 14 \] \[ \text{Imaginary terms: } -3i + 8i = 5i \] \[ \boxed{\,14 + 5i\,} \]

3) Conjugate — product is real

Q. Let \( z=-2+7i \). Show that \( z\overline z \) is real and find its value.

Show solution

\[ \overline z=-2-7i \] \[ z\overline z = (-2)^2 + 7^2 = 4 + 49 = 53 \] \[ \boxed{\,53\in\mathbb{R}\,} \]

4) Division (multiply by the conjugate)

Q. Write \( \dfrac{8-2i}{3+i} \) in the form \( a+bi \).

Show solution

Conjugate step: multiply top and bottom by \(3-i\):

\[ \frac{8-2i}{3+i}\cdot\frac{3-i}{3-i} =\frac{(8-2i)(3-i)}{(3+i)(3-i)}. \]

Numerator area model \((8-2i)(3-i)\)

	\(3\)	\(-i\)
\(8\)	\(24\)	\(-8i\)
\(-2i\)	\(-6i\)	\(2i^2\)

Denominator area model \((3+i)(3-i)\)

	\(3\)	\(-i\)
\(3\)	\(9\)	\(-3i\)
\(i\)	\(3i\)	\(-\,i^2\)

\[ \text{Numerator: } 24-8i-6i+2i^2=22-14i \] \[ \text{Denominator: } 9+(-i^2)=10 \] \[ \frac{8-2i}{3+i}=\frac{22-14i}{10} =\frac{11}{5}-\frac{7}{5}i \] \[ \boxed{\,\tfrac{11}{5}-\tfrac{7}{5}i\,} \]

5) Modulus

Q. Find \( |\, -6+8i \,| \).

Show solution

\[ |z|=\sqrt{a^{2}+b^{2}} \] \[ |\, -6+8i \,|=\sqrt{(-6)^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10 \] \[ \boxed{\,10\,} \]

6) Quadratic with complex roots

Q. Solve \( x^{2}+6x+25=0 \).

Show solution

\[ \textbf{Quadratic formula:}\quad x=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a} \] \[ \textbf{Substitute:}\ a=1,\ b=6,\ c=25 \] \[ x=\dfrac{-6\pm\sqrt{\,6^{2}-4(1)(25)\,}}{2(1)} \] \[ =\dfrac{-6\pm\sqrt{-64}}{2} \] \[ \boxed{\,x=-3\pm4i\,} \]

7) Equating real & imaginary parts

Q. If \( z=a+bi \) satisfies \( 3z-(2-i)=10+5i \), find \( a \) and \( b \).

Show solution

\[ 3(a+bi)-(2-i)=(3a-2)+(3b+1)i \] \[ \text{Real: } 3a-2=10 \ \Rightarrow\ a=4 \] \[ \text{Imaginary: } 3b+1=5 \ \Rightarrow\ b=\dfrac{4}{3} \] \[ \boxed{\,z=4+\dfrac{4}{3}i\,} \]

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