Introduction to Differential Equations,Exercise 1.1,1.5,1.6,1.8,1.9,1.10

As noted,if $z=x+iy$,$x,y\in\mathbf{R}$,then $|z|=\sqrt{x^2+y^2}$ is equivalent to $|z|^2=z\overline{z}$.Use this to show that if also $w\in\mathbf{C}$,
$$
|zw|=|z|\cdot|w|.
$$

Solve:
  $|zw|^{2}=(zw)\cdot
  (\overline{zw})=(zw)\cdot(\overline{z}\cdot\overline{w})=(z\cdot \overline{z})\cdot(w\cdot\overline{w})=|z|^{2}|w|^2$.

Note that
\begin{align*}
  |z+w|^2&=(z+w)(\overline{z}+\overline{w})
\\&=|z|^2+|w|^2+w\overline{z}+z\overline{w}
\\&=|z|^2+|w|^2+2\mathbf{Re}zw.
\end{align*}
Show that $\mathbf{Re}(zw)\leq |zw|$ and use this in concert with an expansion of $(|z|+|w|)^2$ and the first identity above to deduce that
$$
|z+w|\leq |z|+|w|.
$$

---

Evaluate
$$
\int_0^y \frac{dx}{1+x^2}.
$$

solve:Let $x=\tan\theta$.Then
$$
\int_0^y \frac{dx}{1+x^2}=\int_0^y\cos^2\theta dx=\int_0^{\arctan y}\cos^2\theta \frac{d\theta}{\cos^2\theta}=\arctan y.
$$

---

Evaluate
$$
\int_0^y \frac{dx}{\sqrt{1-x^2}}.
$$
Solve:Let $x=\cos t$,where $t\in [0,\pi]$.Then

$$\int_0^y \frac{dx}{\sin t}=\int_{\frac{\pi}{2}}^{\arccos y}-1dt=\frac{\pi}{2}-\arccos y.$$

1.8 Set
$$
\cosh t=\frac{1}{2}(e^t+e^{-t}),\sinh t=\frac{1}{2}(e^t-e^{-t}).
$$
Show that
$$
\frac{d}{dt}\cosh t=\sinh t,\frac{d}{dt}\sinh t=\cosh t,
$$
and
$$
\cosh^2t-\sinh^2t=1.
$$

Proof:Simple.

1.9 Evaluate
$$
\int_0^y \frac{dx}{\sqrt{1+x^2}}.
$$

Solve:Let $x=\sinh t$,so
$$
\int_0^y \frac{dx}{\cosh t}=\int_0^{\sinh^{-1} y}1 dt=\ln (y+\sqrt{1+y^2}).
$$

1.10 Evaluate
$$
\int_0^y \sqrt{1+x^2}dx.
$$

Solve:Let $x=\sinh t$,then
$$
\int_0^y \sqrt{1+x^2}dx=\int_0^{\ln (y+\sqrt{1+y^2})} \cosh^{2} t
dt=\frac{1}{2}\int_0^{\ln(y+\sqrt{1+y^2})}(\cosh 2t+1)dt=\frac{y
  \sqrt{1+y^2}+\ln (y+\sqrt{1+y^2})}{2}.
$$