IB Maths AI 1.9 Notes

The imaginary unit

Complex numbers extend the real number system by introducing the number $i$, where $i^2=-1$. This allows us to solve equations such as $x^2+1=0$, which have no real solutions.

A complex number is written in Cartesian form as $z=a+bi$, where $a,b\in\mathbb{R}$. Here, $a$ is the real part and $b$ is the imaginary part. 

The key fact here is:

$i^2=-1$

From this:

• $i^3=i^2\cdot i=-i$
• $i^4=1$
• powers of $i$ repeat every $4$ steps

Write the following in terms of $i$.

(a) $(2i)^3$

(b) $-\sqrt{-81}$

For (a), $(2i)^3=8i^3=8(-i)=-8i$.

For (b), using the principal square root, $\sqrt{-81}=9i$, so $-\sqrt{-81}=-9i$.

We can now solve equations whose solutions are not real.

Solve:

(a) $5x^2+7=0$

(b) $x^2+x+1=0$

For (a):

$5x^2=-7$

$x^2=-\frac{7}{5}$

$x=\pm\sqrt{-\frac{7}{5}}=\pm i\sqrt{\frac{7}{5}}$

For (b), using the quadratic formula:

$x=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}$