1.12 Complex numbers: Cartesian formIB Maths: Applications and Interpretation HL: Revision notes

There is no real number whose square is negative, so we define $i$ with $i^{2}=-1$. A complex number in Cartesian form is $z=a+bi$ with $a,b\in\mathbb R$. Here $a$ is the real part, $\operatorname{Re}(z)$, and $b$ is the imaginary part, $\operatorname{Im}(z)$. Note that the imaginary part is the real number $b$, not $bi$. Two complex numbers are equal only if both parts are equal.

The conjugate of $z=a+bi$ is $z^{*}=a-bi$. The modulus is the distance from the origin: $|z|=\sqrt{a^{2}+b^{2}}$. For $z=3+4i$: $z^{*}=3-4i$ and $|z|=5$. The argument $\arg z$ is the angle between the positive real axis and the line from the origin to $z$, measured anticlockwise. For $2+3i$, $\arg z=\tan^{-1}\left(\frac32\right)=0.983$ radians. Draw a quick sketch so you place the angle in the correct quadrant. Also $zz^{*}=a^{2}+b^{2}=|z|^{2}$, which is always real.

Add and subtract real and imaginary parts separately: $(3+4i)+(1-2i)=4+2i$. Multiply by expanding and replacing $i^{2}$ with $-1$: $(2-3i)(4+i)=8+2i-12i-3i^{2}=11-10i$. Divide by multiplying top and bottom by the conjugate of the denominator: $\frac{3+4i}{1-2i}=\frac{(3+4i)(1+2i)}{(1-2i)(1+2i)}=\frac{-5+10i}{5}=-1+2i.$ A GDC in complex mode does all of these, and is the way to find powers: $(2-3i)^{4}=-119+120i$. Do small calculations by hand when asked, and check them with the GDC.

On an Argand diagram the horizontal axis is the real axis and the vertical axis is the imaginary axis. The number $a+bi$ is the point $(a,b)$ in the complex plane. The conjugate $a-bi$ is the reflection in the real axis. The modulus is the length of the line from the origin and the argument is its angle. Sums can be drawn by adding the two position vectors. So $3+4i$ is the point $(3,4)$ and $3-4i$ is $(3,-4)$.

For $ax^{2}+bx+c=0$ with real coefficients, the discriminant is $\Delta=b^{2}-4ac$. If $\Delta<0$ there are no real solutions, but the quadratic formula still works with $\sqrt{-k}=i\sqrt{k}$: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$ Example: $x^{2}-6x+25=0$ has $\Delta=-64$, so $x=\frac{6\pm8i}{2}=3\pm4i$. The two roots are complex conjugates.

If $\Delta<0$ the parabola $y=ax^{2}+bx+c$ never meets the $x$-axis, which is why there are no real roots. For $f(x)=x^{2}-4x+13=(x-2)^{2}+9$ the vertex is $(2,9)$, above the axis, and the roots are $2\pm3i$. The real part of the roots, $2$, equals $-\frac{b}{2a}$, the $x$-coordinate of the vertex. The sum of the roots is $-\frac ba=4$ and their product is $\frac ca=13$.