The four operations

Complex numbers, grade 12 (5 units) Lesson 2 of 18Learning

The four operations

Adding and subtracting

Add and subtract each part on its own – the real parts and the imaginary parts:

\displaystyle (a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i

$\displaystyle {(3+2i)+(1-5i)}={4-3i}$ .
$\displaystyle {(3+2i)-(1-5i)}={2+7i}$ .

Multiplying

Multiply as you multiply brackets, and replace $\displaystyle i^2$ by $\displaystyle -1$ :

\displaystyle (a+bi)(c+di)

\displaystyle =

\displaystyle (ac-bd)+(ad+bc)i

The example below multiplies $\displaystyle (2+3i)(1-i)$ :

$\displaystyle {(2+3i)(1-i)}={2-2i+3i-3i^2}$ .
$\displaystyle {-3i^2}={+3}$ , so the result is $\displaystyle 5+i$ .

The short multiplication formulas hold too: $\displaystyle {(1+i)^2}={1+2i+i^2}={2i}$ .

The conjugate

The conjugate of $\displaystyle {z}={a+bi}$ is $\displaystyle {\bar{z}}={a-bi}$ : the same real part, and the imaginary part with the opposite sign. A number times its conjugate is a real number that is not negative:

\displaystyle (a+bi)(a-bi)

\displaystyle =

\displaystyle a^{2}+b^{2}

$\displaystyle {(3+4i)(3-4i)}={9-16i^2}={9+16}={25}$ .
The conjugate of $\displaystyle -2i$ is $\displaystyle 2i$ , and the conjugate of a real number is the number itself.

Dividing

To divide, multiply the numerator and the denominator by the denominator’s conjugate: the denominator becomes a real number. The example below divides $\displaystyle \frac{4+7i}{2+i}$ :

The denominator’s conjugate is $\displaystyle 2-i$ : $\displaystyle \frac{(4+7i)(2-i)}{(2+i)(2-i)}$ .
The numerator: $\displaystyle {8-4i+14i-7i^2}={15+10i}$ . The denominator: $\displaystyle {2^2+1^2}={5}$ .
$\displaystyle {\frac{15+10i}{5}}={3+2i}$ .

Check: $\displaystyle {(3+2i)(2+i)}={6+3i+4i+2i^2}={4+7i}$ .

The laws of arithmetic and order

Every law of arithmetic of the real numbers holds in the complex numbers: the commutative and associative laws of adding and multiplying, and the distributive law. Every number has an opposite, and every number other than 0 has a reciprocal. So everything you did in algebra – opening brackets, taking out a common factor, the short multiplication formulas – holds here too.

But they have no order: you cannot say that $\displaystyle i$ is greater than 0 or less than 0. The proof below checks both possibilities:

If $\displaystyle i>0$ , multiplying both sides by $\displaystyle i$ keeps the sign: $\displaystyle i\cdot i>0$ , that is $\displaystyle -1>0$ – a contradiction.
If $\displaystyle i<0$ , then $\displaystyle -i>0$ , and again $\displaystyle (-i)(-i)>0$ , that is $\displaystyle -1>0$ – a contradiction.

So $\displaystyle z_1<z_2$ has no meaning in complex numbers.

Exercise 1

Calculate, and write the result in algebraic form.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

Divide: multiply the numerator and the denominator by the denominator’s conjugate.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

Solve the equations. Get $\displaystyle z$ alone, then divide.

Formulas: complex numbers, grade 12 (5 units)

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The algebraic form and the four operations

The number $\displaystyle i$
$\displaystyle i^{2}=-1$
The algebraic form
$\displaystyle z=a+bi$ $\displaystyle a=\operatorname{Re}(z),\ b=\operatorname{Im}(z)$
Equality of complex numbers
$\displaystyle a+bi=c+di\iff a=c,\ b=d$
Multiplying
$\displaystyle (a+bi)(c+di)$ $\displaystyle =$ $\displaystyle (ac-bd)+(ad+bc)i$
The conjugate
$\displaystyle \bar{z}=a-bi$
A number times its conjugate
$\displaystyle z\cdot\bar{z}$ $\displaystyle =$ $\displaystyle a^{2}+b^{2}$
Dividing – times the denominator’s conjugate
$\displaystyle \frac{z_{1}}{z_{2}}$ $\displaystyle =$ $\displaystyle \frac{z_{1}\cdot\bar{z}_{2}}{z_{2}\cdot\bar{z}_{2}}$

The modulus and the conjugate

The modulus – the distance from the origin
$\displaystyle |z|$ $\displaystyle =$ $\displaystyle \sqrt{a^{2}+b^{2}}$
The distance between two points
$\displaystyle |z_{1}-z_{2}|$ $\displaystyle =$ $\displaystyle \sqrt{(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}}$
The product with the conjugate and the modulus
$\displaystyle z\cdot\bar{z}$ $\displaystyle =$ $\displaystyle |z|^{2}$
The modulus of a product
$\displaystyle |z_{1}\cdot z_{2}|$ $\displaystyle =$ $\displaystyle |z_{1}|\cdot|z_{2}|$
The conjugate of a sum
$\displaystyle \overline{z_{1}+z_{2}}$ $\displaystyle =$ $\displaystyle \bar{z}_{1}+\bar{z}_{2}$
The conjugate of a product
$\displaystyle \overline{z_{1}\cdot z_{2}}$ $\displaystyle =$ $\displaystyle \bar{z}_{1}\cdot\bar{z}_{2}$
A number plus its conjugate
$\displaystyle z+\bar{z}$ $\displaystyle =$ $\displaystyle 2\operatorname{Re}(z)$
The triangle inequality
$\displaystyle |z_{1}+z_{2}|\le|z_{1}|+|z_{2}|$

The quadratic equation

The quadratic formula
$\displaystyle z_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
The sum of the roots
$\displaystyle z_{1}+z_{2}$ $\displaystyle =$ $\displaystyle -\frac{b}{a}$
The product of the roots
$\displaystyle z_{1}\cdot z_{2}$ $\displaystyle =$ $\displaystyle \frac{c}{a}$

The polar form (angles in radians)

The polar form
$\displaystyle z=r(\cos\theta+i\sin\theta)$ $\displaystyle r=|z|$
The notation cis
$\displaystyle r\operatorname{cis}\theta$ $\displaystyle =$ $\displaystyle r(\cos\theta+i\sin\theta)$
The real part
$\displaystyle a=r\cos\theta$
The imaginary part
$\displaystyle b=r\sin\theta$
The angle – by the quadrant of the point
$\displaystyle \tan\theta=\frac{b}{a}$ $\displaystyle a\neq0$
A product in polar form
$\displaystyle z_{1}\cdot z_{2}=r_{1}\cdot r_{2}\left[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})\right]$
A quotient in polar form
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}\operatorname{cis}(\theta_{1}-\theta_{2})$
De Moivre’s theorem
$\displaystyle \left[R(\cos\varphi+i\sin\varphi)\right]^{n}$ $\displaystyle =$ $\displaystyle R^{n}(\cos n\varphi+i\sin n\varphi)$
The solutions of $\displaystyle {z^n}={R(\cos\varphi+i\sin\varphi)}$
$\displaystyle z_{k}=\sqrt[n]{R}\left[\cos\left(\frac{\varphi}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\varphi}{n}+\frac{2k\pi}{n}\right)\right]$ $\displaystyle z^{n}=R(\cos\varphi+i\sin\varphi),\ k=0,1,\ldots,n-1$

Loci and rotation

The circle with centre $\displaystyle z_0$ and radius $\displaystyle R$
$\displaystyle |z-z_{0}|=R$
The perpendicular bisector of the segment $\displaystyle z_1z_2$
$\displaystyle |z-z_{1}|=|z-z_{2}|$
Apollonius’ circle
$\displaystyle |z-z_{1}|=m|z-z_{2}|$ $\displaystyle m>0,\ m\neq1$
Rotating by $\displaystyle \theta$ about $\displaystyle z_0$
$\displaystyle w=z_{0}+(z-z_{0})\operatorname{cis}\theta$