Tag Archives: imaginary numbers

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Magnitudes, Modulus, Proofs, Roots, Surds

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

November 4, 2018 Tiago Hands

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.

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Mathematics Proofs, Trigonometric Identities

Proving that arg(z_1/z_2)=arg(z_1)-arg(z_2)

October 28, 2018 Tiago Hands

In this post I’ll be proving to you that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Now firstly I will have to say that:

${ z }_{ 1 }={ r }_{ 1 }\left( \cos { { \theta }_{ 1 }+i\sin { { \theta }_{ 1 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }$

And also that:

${ z }_{ 2 }={ r }_{ 2 }\left( \cos { { \theta }_{ 2 }+i\sin { { \theta }_{ 2 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }$

If this is the case, then…

Since this is in the form:

$z=r\left( \cos { \theta +i\sin { \theta } } \right)$

I would have to conclude that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) ={ \theta }_{ 1 }-{ \theta }_{ 2 }=arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Hence I’ve proven that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Angles, Complex Numbers, Further Pure Maths, Imaginary Numbers, Trigonometric Identities

arg(z_1*z_2)=arg(z_1)+arg(z_2) Proof

October 15, 2018 Tiago Hands

In this post I’ll be proving why:

$arg\left( { z }_{ 1 }{ z }_{ 2 } \right) =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)$

Let’s say that:

${ z }_{ 1 }={ r }_{ 1 }\left( \cos { \left( { \theta }_{ 1 } \right) +i\sin { \left( { \theta }_{ 1 } \right) } } \right)$

And also that:

${ z }_{ 2 }={ r }_{ 2 }\left( \cos { \left( { \theta }_{ 2 } \right) +i\sin { \left( { \theta }_{ 2 } \right) } } \right)$

This would imply that:

$arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }$

$arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }$

Now if we multiply ${ z }_{ 1 }$ and ${ z }_{ 2 }$ together, we get:

Which is thanks to what we know about trigonometric identities.

As we can see above, we’ve formed another complex number:

${ z }_{ 1 }{ z }_{ 2 }={ r }_{ 1 }{ { r }_{ 2 }\left( \cos { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) +i\sin { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) } } \right) }$

And this is in the form of:

$z=r\left( \cos { \left( \theta \right) +i\sin { \left( \theta \right) } } \right)$

And because of the rules of complex numbers, we can say that:

$arg\left( { z }_{ 1 }{ z }_{ 2 } \right) \\ \\ ={ { \theta } }_{ 1 }+{ { \theta } }_{ 2 }\\ \\ =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)$

Hence, we have our proof.

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Mathematics Proofs, Modulus, Pythagoras

How to prove that |z_1||z_2|=|z_1z_2|, Complex Numbers

October 14, 2018 Tiago Hands

In this post I’ll be showing you how to prove that:

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

Firstly, let’s say that:

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

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

If this is the case, then according to the rules of complex numbers:

$\left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

$\left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

Secondly, let’s determine what ${ z }_{ 1 }{ z }_{ 2 }$ is…

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

As you can see, we get the result above – which is another complex number.

This means that:

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

Therefore we’ve proven that:

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

You can watch a video related to this proof below…

Algebra, Complex Numbers, Further Pure Maths, Imaginary Numbers, Magnitudes, Modulus, Pythagoras, Vectors

Further Pure Maths: Complex Number Proof (1)

July 26, 2017 Tiago Hands

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:

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

Maths Proofs, Videos & Tutorials

Tag Archives: imaginary numbers

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

Proving that arg(z_1/z_2)=arg(z_1)-arg(z_2)

arg(z_1*z_2)=arg(z_1)+arg(z_2) Proof

How to prove that |z_1||z_2|=|z_1z_2|, Complex Numbers

Further Pure Maths: Complex Number Proof (1)

GCSE + A Level Mathematics Proofs, Videos and Tutorials.