Problems from Definition of complex numbers

Development:

The first one is a complex number, but it is not a pure imaginary since it has a real part. The pure imaginary number is $$\sqrt{-27}$$. This is because it is a multiple of the imaginary unit.

Solution:

The pure imaginary is $$\sqrt{-27}$$.

Hide solution and development

Development:

If they must have a multiple of the imaginary unit, the easiest thing is to take an imaginary number, equate it to $$x$$ and then square it: $$$12i=x \ \Rightarrow \ (12i)^2=x^2 \ \Rightarrow \ 144i^2=x^2 \ \Rightarrow \ -144=x^2 \ \Rightarrow \ x^2+144=0$$$ it has $$12i$$ as solution, and $$12i$$ is a multiple of the imaginary unit $$i$$. Similarly we can obtain $$x^2+169=0$$ has a multiple of $$i$$. The solution would be $$13i$$.

Solution:

$$ x^2+144=0 \quad $$ and $$ \quad x^2+169=0$$.

Hide solution and development

Development:

1. $$3x^2+27=0 \ \Rightarrow \ 3x^2=-27 \ \Rightarrow \ x=\pm \sqrt{-\dfrac{27}{3}}=\pm 3i$$
2. $$8x^2+4x-2=0 \ \Rightarrow$$ $$ \displaystyle \begin{array}{rl} x &=\frac{-4\pm\sqrt{16-4\cdot8\cdot2}}{16}= \frac{-4\pm\sqrt{-48}}{16}=\frac{-1}{4}\pm \frac{i\sqrt{48}}{16}\\ &=\frac{-1}{4}\pm \frac{i\sqrt{2^4\cdot3}}{16} = \frac{-1}{4}\pm \frac{i\cdot 2^2\sqrt{3}}{16}= \frac{-1}{4}\pm \frac{i\sqrt{3}}{4} \end{array} $$

Solution:

1. $$x=\pm 3i$$
2. $$x_1= \dfrac{-1}{4}+ \dfrac{\sqrt{3}}{4}i \qquad x_2= \dfrac{-1}{4}- \dfrac{\sqrt{3}}{4}i $$

Hide solution and development