# Further Pure Maths: Complex Number Proof (1)

In this post, I’ll be proving that: $\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

First of all, let’s say that:

${ z }_{ 1 }=x+iy$

Whereby, $\left\{ x\in R,\quad y\in R \right\}$.

And also that:

${ z }_{ 2 }=p+iq$

Whereby, $\left\{ p\in R,\quad q\in R \right\}$.

If this is the case, this means that:

${ z }_{ 1 }\cdot { z }_{ 2 }=\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px-qy+i\left( qx+py \right)$

Therefore:

$LHS\\ \\ =\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+\left\{ { q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } \right\} } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \\ \\ =RHS$

Hence we’ve proven that:

$\left| { z }_{ 1 }\cdot { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

# Vector Magnitude Proof

Want to find the magnitude of a vector?

You quite simply have to know that:

${ \left| \underline { v } \right| }^{ 2 }={ \underline { v } }^{ 2 }$

And that if:

$\underline { v } =\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) ,\quad { \underline { v } }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$

So:

$If\quad { \underline { v } }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 },\\ \\ \therefore \quad { \left| \underline { v } \right| }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \therefore \quad \left| \underline { v } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \\$