Aims for foundations and connections: To improve the ability of the students to read and write mathematics well; to enable them to develop effective study skills for individual and group working; to introduce students to the beauty, the ubiquity and the importance of university-level mathematics.

这是一份Bath巴斯大学MA10265R作业代写的成功案

The second inequality is of the form $t^{2}>A$ and, as in Example 6 , a numerical illustration will help. If we have $t^{2}>4$ then $t$ could be positive or negative, giving two possibilities, namely $t>2$ or $t<-2$. So Case 2 becomes $x<0$ and $\left(x<\frac{3}{2}-\frac{\sqrt{13}}{2}\right.$ or $\left.x>\frac{3}{2}+\frac{\sqrt{13}}{2}\right) .$
The fact that $x$ is negative means that the third inequality cannot be satisfied, and so this case reduces to
$$
x<\frac{3}{2}-\frac{\sqrt{13}}{2} .
$$
Finally, putting together Cases 1 and 2 gives the complete solution as
$$
x<\frac{3}{2}-\frac{\sqrt{13}}{2} \text { or } 0<x<\frac{3}{2}+\frac{\sqrt{13}}{2} \text {. }
$$

MA10274 COURSE NOTES ：

From the definition of the modulus we can calculate that $\left|z_{1}\right|=2$ and $\left|z_{2}\right|=\sqrt{2}$.
From the diagram we can find the angles, so $\theta_{1}=120^{\circ}=2 \pi / 3$ and $\theta_{2}=135^{\circ}=3 \pi / 4$. This gives all the information we need, and so
$$
z_{1}=2\left(\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}\right), \quad z_{2}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+\mathrm{i} \sin \frac{3 \pi}{4}\right) .
$$
If we now use the polar product formula we obtain
$$
z_{1} z_{2}=2 \sqrt{2}\left(\cos \frac{17 \pi}{12}+\mathrm{i} \sin \frac{17 \pi}{12}\right) .
$$