As noted,if $z=x+iy$,$x,yinmathbf{R}$,then $|z|=sqrt{x^2+y^2}$ is equivalent to $|z|^2=zoverline{z}$.Use this to show that if also $winmathbf{C}$,
$$
|zw|=|z|cdot|w|.
$$
Solve:
$|zw|^{2}=(zw)cdot
(overline{zw})=(zw)cdot(overline{z}cdotoverline{w})=(zcdot overline{z})cdot(wcdotoverline{w})=|z|^{2}|w|^2$.
Note that
egin{align*}
|z+w|^2&=(z+w)(overline{z}+overline{w})
\&=|z|^2+|w|^2+woverline{z}+zoverline{w}
\&=|z|^2+|w|^2+2mathbf{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 $x= an heta$.Then
$$
int_0^y frac{dx}{1+x^2}=int_0^ycos^2 heta dx=int_0^{arctan y}cos^2 heta frac{d heta}{cos^2 heta}=arctan y.
$$
Evaluate
$$
int_0^y frac{dx}{sqrt{1-x^2}}.
$$
Solve:Let $x=cos t$,where $tin [0,pi]$.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
$$
int_0^y sqrt{1+x^2}dx=int_0^{ln (y+sqrt{1+y^2})} cosh^{2} t
dt=frac{1}{2}int_0^{ln(y+sqrt{1+y^2})}(cosh 2t+1)dt=frac{y
sqrt{1+y^2}+ln (y+sqrt{1+y^2})}{2}.
$$