# |z_1|/|z_2|=|z_1/z_2| proof (algebraic)

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

Firstly, I’ll say that: ${ z }_{ 1 }=x+iy,\quad \therefore \quad \left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

And also that: ${ z }_{ 2 }=p+iq,\quad \therefore \quad \left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

If this is the case, then: $\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { x+iy }{ p+iq } \\ \\ =\frac { \left( x+iy \right) }{ \left( p+iq \right) } \cdot \frac { \left( p-iq \right) }{ \left( p-iq \right) } \\ \\ =\frac { px-iqx+ipy-{ i }^{ 2 }qy }{ { p }^{ 2 }-ipq+ipq-{ i }^{ 2 }{ q }^{ 2 } } \\ \\ =\frac { \left( px+qy \right) +i\left( py-qx \right) }{ { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) +i\left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right)$

And as this is in the form: $z=a+ib$

I would have to conclude that: $RHS=\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| \\ \\ =\sqrt { { \left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 }+{ \left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } +\frac { { \left( py-qx \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 }+{ \left( py-qx \right) }^{ 2 } }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) ^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }-2pqxy+{ q }^{ 2 }{ x }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } } \\ \\ =\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =LHS$

Hence, I have my proof.

# More ways in which to express the golden ratio

In this blog post I’ll be revealing more ways (4 in fact) in which to express or come up with the value of the golden ratio

Number One: $\varphi =\frac { a+b }{ a } \\ \\ =\frac { a }{ a } +\frac { b }{ a } \\ \\ =1+{ \left( \varphi \right) }^{ -1 }\\ \\ =1+\frac { 1 }{ \varphi }$

Number Two: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad { \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad { \varphi }^{ 2 }-\varphi =1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }=1\\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 4 }{ 4 } +\frac { 1 }{ 4 } \\ \\ \Rightarrow \quad { \left( \varphi -\frac { 1 }{ 2 } \right) }^{ 2 }=\frac { 5 }{ 4 } \\ \\ \Rightarrow \quad \varphi -\frac { 1 }{ 2 } =\frac { \sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad \varphi =\frac { 1 }{ 2 } +\frac { \sqrt { 5 } }{ 2 } \\ \\ \therefore \quad \varphi =\frac { 1+\sqrt { 5 } }{ 2 }$

Number Three: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } \\ \\ \Rightarrow \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ \varphi } } } \\ \\ \therefore \quad \varphi =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

Number Four: $\varphi =1+\frac { 1 }{ \varphi } \\ \\ { \Rightarrow \quad \varphi }^{ 2 }=\varphi +1\\ \\ \Rightarrow \quad \varphi =\sqrt { \varphi +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \varphi +1 } +1 } \\ \\ \Rightarrow \quad \varphi =\sqrt { \sqrt { \sqrt { \varphi +1 } +1 } +1 } \\ \\ \therefore \quad \varphi =\sqrt { \sqrt { \sqrt { \frac { 1+\sqrt { 5 } }{ 2 } +1 } +1 } +1 } \\ \\$

And check out this calculator trick…

If you’re not satisfied with what I’ve already produced, then you can have a go at proving that… $\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } } \\ \\$

Without using the phi (φ) symbol.

Enjoy!!! 😀

# Another way to express the golden ratio mathematically

In this post I’m going to be proving that… $\varphi =\frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

So, here I go… $x=\frac { 1+\sqrt { 5 } }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 1+2\sqrt { 5 } +5 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 4 }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2+2\sqrt { 5 } }{ 4 } \\ \\ \Rightarrow \quad { x }^{ 2 }-1=\frac { 2 }{ 2 } \cdot \frac { \left( 1+\sqrt { 5 } \right) }{ 2 }$

Wait for it… $\Rightarrow \quad { x }^{ 2 }-1=1\cdot x\\ \\ \Rightarrow \quad { x }^{ 2 }-1=x\\ \\ \Rightarrow \quad { x }^{ 2 }=x+1\\ \\ \Rightarrow \quad \frac { { x }^{ 2 } }{ x } =\frac { x }{ x } +\frac { 1 }{ x } \\ \\ \Rightarrow \quad x=1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ x } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ x } } \\ \\ \Rightarrow \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ \left( 1+\frac { 1 }{ x } \right) } } \\ \\ \therefore \quad \frac { 1+\sqrt { 5 } }{ 2 } =1+\frac { 1 }{ 1+\frac { 1 }{ 1+\frac { 1 }{ 1+... } } }$

This expression for the golden ratio is quite common, however, before I produced this post – I think it would’ve been very hard to figure out how to derive it from scratch. There aren’t many quirky proofs like this one on the internet – I am quite certain. I hope you liked reading this post! 😀

# Completing The Square (Why It Works)

Prove that: ${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }-bx$

Proof: ${ \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }-bx$

Second Proof: $As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x-\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x-\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x-\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x-\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =x\left( x-2\frac { b }{ 2 } \right) \\ \\ =x\left( x-b \right) \\ \\ ={ x }^{ 2 }-bx$

Prove that: ${ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }={ x }^{ 2 }+bx$

Proof: ${ \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } \right) -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+2x\frac { b }{ 2 } +\frac { { b }^{ 2 } }{ 4 } -\frac { { b }^{ 2 } }{ 4 } \\ \\ ={ x }^{ 2 }+bx$

Second Proof: $As:\\ \\ { p }^{ 2 }-{ q }^{ 2 }=\left( p+q \right) \left( p-q \right) \\ \\ And:\\ \\ p=\left( x+\frac { b }{ 2 } \right) \\ \\ And:\\ \\ q=\frac { b }{ 2 } ,\\ \\ { \left( x+\frac { b }{ 2 } \right) }^{ 2 }-{ \left( \frac { b }{ 2 } \right) }^{ 2 }\\ \\ =\left( x+\frac { b }{ 2 } +\frac { b }{ 2 } \right) \left( x+\frac { b }{ 2 } -\frac { b }{ 2 } \right) \\ \\ =\left( x+2\frac { b }{ 2 } \right) x\\ \\ =x\left( x+b \right) \\ \\ ={ x }^{ 2 }+bx$

# How To Derive The Quadratic Formula $a{ x }^{ 2 }+bx+c=0\\ \\ a{ x }^{ 2 }+bx=-c\\ \\ { x }^{ 2 }+\frac { b }{ a } x=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-{ \left( \frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } =-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { 4ac }{ 4{ a }^{ 2 } } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } \\ \\ x+\frac { b }{ 2a } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ \sqrt { 4 } \sqrt { { a }^{ 2 } } } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=-\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

— USEFUL FORMULAS: $\frac { a }{ c } \pm \frac { b }{ c } =\frac { a\pm b }{ c } \\ \\ \frac { a }{ b } \pm \frac { c }{ d } =\frac { ad }{ bd } \pm \frac { bc }{ bd } =\frac { ad\pm bc }{ bd } \\ \\ \sqrt { { a }^{ 2 } } ={ a }^{ \frac { 2 }{ 2 } }=a\\ \\ \sqrt { a } \sqrt { b } =\sqrt { ab } \\ \\ \frac { \sqrt { a } }{ \sqrt { b } } =\sqrt { \frac { a }{ b } }$