Real number(R), Complex number(C)

Question

計算分數的複數
\(a,b\in R\)
\(z=a+bi\), \(i=\sqrt[]{-1}\)
Real part \(\Rightarrow\) \(Re(z)=a\)
Imaginary part \(\Rightarrow\) \(Im(z)=b\)
\(R\in C\Rightarrow r+0i, r\in R\)
\(z=a+bi, z\in C\)
Define conjugate: \(a-bi=\overline{z}\)
Suppose \(z=a+bi, w=c+di\Rightarrow\overline{z}=a-bi,\overline{w}=c-di\)
\(\overline{z+w}=\overline{z}+\overline{w}\)
proof:
\(z+w=(a+bi)+(c+di)=(a+c)+(b+d)i\)
\(\Rightarrow\overline{z+w}=(a+c)-(b+d)i\)
\(=(a-bi)+(c-di)=\overline{z}+\overline{w}.\)
\(\overline{z\cdot w}=\overline{z}\cdot\overline{w}\)
proof:
\(z\cdot w=ac+adi+bci-bd\)
\(=(ac-bd)+(ad+bc)i\)
\(\Rightarrow \overline{z\cdot w}=(ac-bd)-(ad+bc)i\)
\(\overline{z}\cdot\overline{w}=ac-adi-bci-bd\)
\(=(ac-bd)-(ad+bc)i = \overline{z\cdot w}\)
\(\overline{z}=z\Leftrightarrow (z\in R) \wedge (z-R=0i)\)
proof: \(a-bi=a+bi\Leftrightarrow b=0\)

Absolute value

Suppose \(z=a+bi\in C\)
\(|z|=\sqrt[]{a^2+b^2}\)
\(z\overline{z}=(a+bi)(a-bi)=a^2+b^2=|z|^2\)
\(z+\overline{z}=(a+bi)+(a-bi)=2a=2Re(z)\)
\(\frac{1}{z}=\frac{1}{a+bi} =\frac{a-bi}{(a+bi)(a-bi)} =\frac{a-bi}{a^2+b^2}=\frac{\overline{z}}{|z|^2} \)
Suppose \(z,w\in C\)
\(|zw|=|z||w|\)
proof:
\(|zw|^2=(zw)(\overline{zw})=(z\overline{z})(w\overline{w}) =|z|^2|w|^2 \Rightarrow |zw|=|z||w| \)
\(|z+w|\leq |z|+|w| \)
proof:
\( |z+w|^2=(z+w)\overline{(z+w)} \)
\( =(z+w)(\overline{z}+\overline{w}) \)
\( =z\overline{z}+z\overline{w}+w\overline{z}+w\overline{w} \)
\( =|z|^2+(z\overline{w}+(\overline{z\overline{w}}))+|w|^2 \)
\( =|z|^2+2Re(z\overline{w})+|w|^2 \)
\( \leq |z|^2 + 2|z\overline{w}|+|w|^2 \)
\( =|z|^2+2|z||\overline{w}|+|w|^2 \)
\( =(|z|+|w|)^2 \)
\( \Rightarrow |z+w| \leq |z| + |w| \)

Reference

黃子嘉-線性代數