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

# Mathematical Art Work, including Isometric Drawings (Visualising Maths)

Over the past couple of months I’ve been drawing mathematical figures and logos. Quite recently I discovered that art can help me understand complex mathematics a little better, especially three dimensional problems. It turns out that infinitely many complex sketches can emerge out of isometric fields which can be created using pencils, rulers and compasses. Isometric fields are used by engineers and architects who model three dimensional structures. Isometric fields allow you to represent cubes and various different 3 dimensional shapes on 2 dimensional surfaces in quite spectacular fashion. Here are some pieces of work I produced using such fields…  Follow Mathematics Videos & Proofs’s board Mathematical Art & Beauty on Pinterest.

If you’d like to see more work that I’ve been able to produce, please visit my mathematics Pinterest page. I’ve thoroughly enjoyed drawing optical illusions and logos using isometric paper, and I could indeed start designing isometric logos and pieces of text to make some extra money whilst studying mathematics. It’s not that often you bump into sound business ideas, especially business ideas related to both mathematics and art.

If you are a mathematics student but have never used isometric paper before, I’d recommend downloading isometric paper via the links below and producing mathematical sketches. Isometric paper could potentially be very useful to those interested in learning more about vector geometry and extra dimensions.

And by the way, thanks for stopping by. Enjoy your new year celebrations! 🙂

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

# Vector Proof – Angle Between Two Vectors

Prove that: $cos\theta =\frac { \underline { a } \cdot \underline { b } }{ \left| \underline { a } \right| \left| \underline { b } \right| }$

Firstly, look at the image below: Also know that: $cosC=\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab }$ $\left| \underline { a } \right| \left| \underline { a } \right| ={ \underline { a } }^{ 2 }\\ \\ \left| \underline { b } \right| \left| \underline { b } \right| ={ \underline { b } }^{ 2 }\\ \\ \left| \underline { b } -\underline { a } \right| \left| \underline { b } -\underline { a } \right| ={ \left( \underline { b } -\underline { a } \right) }^{ 2 }$

From the image, you’ll be able to see that: $a=\left| \underline { b } \right| \\ \\ b=\left| \underline { a } \right| \\ \\ c=\left| \underline { b } -\underline { a } \right| \\ \\ cosC=cos\theta ,\quad \therefore \quad C=\theta$

Now: # Vector Proof (1)

Prove that: $\left( \begin{matrix} { a }_{ 1 } \\ { a }_{ 2 } \\ { a }_{ 3 } \end{matrix} \right) \left( \begin{matrix} { b }_{ 1 } \\ { b }_{ 2 } \\ { b }_{ 3 } \end{matrix} \right) ={ a }_{ 1 }{ b }_{ 1 }+{ a }_{ 2 }{ b }_{ 2 }+{ a }_{ 3 }{ b }_{ 3 }\\ \\$

Firstly, look at the image below. You should know that, if $\underline { a } \\ \\$ and $\underline { b } \\ \\$ are perpendicular $\underline { a } \cdot \underline { b } =0\\ \\$.

You should also know these rules: $\left| \underline { i } \right| \left| \underline { i } \right| ={ \underline { i } }^{ 2 }=1\cdot 1=1\\ \\ \left| \underline { j } \right| \left| \underline { j } \right| ={ \underline { j } }^{ 2 }=1\cdot 1=1\\ \\ \left| \underline { k } \right| \left| \underline { k } \right| ={ \underline { k } }^{ 2 }=1\cdot 1=1\\ \\$

Knowing these rules, we can say that: *Click on the proof above to see it in full.

# FREE MATHS VIDEOS – FOR A LEVEL & GCSE STUDENTS

Welcome to MathsVideos.net. On this website you will find maths videos containing tricks and advice that you can use to pass your maths exams. MathsVideos.net was recently created in January 2014. These videos are produced by a student who is doing his A-Levels in the UK.