Mathematics 9709 · Pure Mathematics 3

Complex arithmetic and the conjugate

A complex number $z=x+iy$ is two real numbers joined by $i$ . Add and multiply just like brackets, then use $i^2=-1$ to tidy up.
Only division is new: multiply the top and the bottom by the conjugate $z^*=x-iy$ . This makes the bottom a plain real number.

To divide, multiply the top and the bottom by the conjugate of the bottom. It works because $z\,z^*=(x+iy)(x-iy)=x^2+y^2$ is always real. It is the same move you used to clear a surd from the bottom of a fraction.

Add, multiply, and divide by the conjugate

Add / subtract: put real with real, imaginary with imaginary.
Multiply: expand the brackets, then swap $i^2$ for $-1$ .
Divide: clear the bottom using its conjugate:

$z^*=x-iy,\qquad z\,z^*=x^2+y^2,\qquad \frac{1}{z}=\frac{z^*}{x^2+y^2}$

Two complex numbers are equal only when the real parts match and the imaginary parts match.
So one complex equation gives you two real equations to solve together.

On the Argand diagram, z* is z flipped across the real axis.

A complex number works like a 2D vector

Compare $z=x+iy$ with the column $\begin{pmatrix}x\\y\end{pmatrix}$ from the Vectors chapter. They hold the same two numbers. Adding two complex numbers joins the arrows tip-to-tail. $|z|$ is the arrow's length, and $\arg z$ is its direction. That is why the next topics look like geometry: taking the conjugate flips the arrow across the real axis, and multiplying turns it and changes its length.

Worked examples

Worked example

Multiply $(2+3i)(8-i)$

Expand all four products: $(2)(8)+(2)(-i)+(3i)(8)+(3i)(-i)=16-2i+24i-3i^2$ .
Replace $i^2$ with $-1$ : $16+22i-3(-1)=16+22i+3$ .

Answer

19+22i

Worked example

Real question: $u=-4+2i$ , $v=3+i$ , find $u/v$ in the form $x+iy$

9709/32 Feb/March 2021 Q8(a).

Multiply top and bottom by the conjugate of the denominator, $3-i$ : $\dfrac{-4+2i}{3+i}\cdot\dfrac{3-i}{3-i}$ .
Bottom becomes real: $(3+i)(3-i)=3^2+1^2=10$ .
Top: $(-4+2i)(3-i)=-12+4i+6i-2i^2=-12+10i+2=-10+10i$ .
Divide each part by 10: $\dfrac{-10+10i}{10}$ .

Answer

u/v=-1+i

Worked example

Equate parts: find real $a$ , $b$ with $(a+bi)(1+2i)=3+4i$

Expand: $(a-2b)+(2a+b)i=3+4i$ (using $i^2=-1$ ).
Match parts: $a-2b=3$ and $2a+b=4$ .
Solve the pair: $a=\tfrac{11}{5},\ b=-\tfrac{2}{5}$ .

Answer

a=\tfrac{11}{5},\quad b=-\tfrac{2}{5}

Where the marks go

Common mistake

Multiply by the conjugate of the bottom (flip only the imaginary sign), top and bottom. Using the wrong conjugate, or only doing the bottom, gives the wrong answer.

Common mistake

When two complex numbers are equal, match the real parts and match the imaginary parts on their own. That turns one equation into a pair you can solve. Never set a real part equal to an imaginary part.

Now you try

Find $u$ if $u(3-5i)=-1+11i$ . Give it as $x+iy$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Move the bracket to the bottom, then multiply by the conjugate: $u=\dfrac{-1+11i}{3-5i}\cdot\dfrac{3+5i}{3+5i}=\dfrac{-58+28i}{34}$ :

Answer

u=-\tfrac{29}{17}+\tfrac{14}{17}i

PreviousImaginary numbers NextThe Argand diagram: modulus and argument

Complex arithmetic and the conjugate

A complex number $z=x+iy$ is two real numbers joined by $i$ . Add and multiply just like brackets, then use $i^2=-1$ to tidy up.
Only division is new: multiply the top and the bottom by the conjugate $z^*=x-iy$ . This makes the bottom a plain real number.

Add, multiply, and divide by the conjugate

Add / subtract: put real with real, imaginary with imaginary.
Multiply: expand the brackets, then swap $i^2$ for $-1$ .
Divide: clear the bottom using its conjugate:

$z^*=x-iy,\qquad z\,z^*=x^2+y^2,\qquad \frac{1}{z}=\frac{z^*}{x^2+y^2}$

Two complex numbers are equal only when the real parts match and the imaginary parts match.
So one complex equation gives you two real equations to solve together.

On the Argand diagram, z* is z flipped across the real axis.

A complex number works like a 2D vector

Worked examples

Worked example

Multiply $(2+3i)(8-i)$

Expand all four products: $(2)(8)+(2)(-i)+(3i)(8)+(3i)(-i)=16-2i+24i-3i^2$ .
Replace $i^2$ with $-1$ : $16+22i-3(-1)=16+22i+3$ .

Answer

19+22i

Worked example

Real question: $u=-4+2i$ , $v=3+i$ , find $u/v$ in the form $x+iy$

9709/32 Feb/March 2021 Q8(a).

Multiply top and bottom by the conjugate of the denominator, $3-i$ : $\dfrac{-4+2i}{3+i}\cdot\dfrac{3-i}{3-i}$ .
Bottom becomes real: $(3+i)(3-i)=3^2+1^2=10$ .
Top: $(-4+2i)(3-i)=-12+4i+6i-2i^2=-12+10i+2=-10+10i$ .
Divide each part by 10: $\dfrac{-10+10i}{10}$ .

Answer

u/v=-1+i

Worked example

Equate parts: find real $a$ , $b$ with $(a+bi)(1+2i)=3+4i$

Expand: $(a-2b)+(2a+b)i=3+4i$ (using $i^2=-1$ ).
Match parts: $a-2b=3$ and $2a+b=4$ .
Solve the pair: $a=\tfrac{11}{5},\ b=-\tfrac{2}{5}$ .

Answer

a=\tfrac{11}{5},\quad b=-\tfrac{2}{5}

Where the marks go

Common mistake

Multiply by the conjugate of the bottom (flip only the imaginary sign), top and bottom. Using the wrong conjugate, or only doing the bottom, gives the wrong answer.

Common mistake

Now you try

Find $u$ if $u(3-5i)=-1+11i$ . Give it as $x+iy$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Move the bracket to the bottom, then multiply by the conjugate: $u=\dfrac{-1+11i}{3-5i}\cdot\dfrac{3+5i}{3+5i}=\dfrac{-58+28i}{34}$ :

Answer

u=-\tfrac{29}{17}+\tfrac{14}{17}i