From c526004533c1b7a3bb623e4caf1ec609b623e1e4 Mon Sep 17 00:00:00 2001 From: Andrew Lorimer Date: Wed, 13 Feb 2019 08:45:12 +1100 Subject: [PATCH] fix formatting for complex/imaginary notes --- spec/complex.md | 28 +++++++++++----------------- 1 file changed, 11 insertions(+), 17 deletions(-) diff --git a/spec/complex.md b/spec/complex.md index 5809170..374b793 100755 --- a/spec/complex.md +++ b/spec/complex.md @@ -8,9 +8,8 @@ $\therefore i = \sqrt {-1}$ ### Simplifying negative surds -$\sqrt{-2} = \sqrt{-1 \times 2}$ - -          $= \sqrt{2}i$ +$\sqrt{-2} = \sqrt{-1 \times 2}$ +$= \sqrt{2}i$ ## Complex numbers @@ -23,21 +22,16 @@ General form: $z=a+bi$ ### Addition -If $z_1 = a+bi$ and $z_2=c+di$, then - -            $z_1+z_2 = (a+c)+(b+d)i$ +If $z_1 = a+bi$ and $z_2=c+di$, then +$z_1+z_2 = (a+c)+(b+d)i$ ### Subtraction -If $z_1=a+bi$ and $z_2=c+di$, then - -           $z_1−z_2=(a−c)+(b−d)i$ +If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$ ### Multiplication by a real constant -If $z=a+bi$ and $k \in \mathbb{R}$, then - -           $kz=ka+kbi$ +If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$ ### Powers of $i$ $i^0=1$ @@ -48,17 +42,18 @@ $i^4=1$ $\dots$ Therefore.. + - $i^{4n} = 1$ - $i^{4n+1} = i$ - $i^{4n+2} = -1$ - $i^{4n+3} = -i$ -Divide by 4 and take remainder +Divide by 4 and take remainder. ### Multiplying complex expressions -If $z_1 = a+bi$ and $z_2=c+di$, then -           $z_1 \times z_2 = (ac-bd)+(ad+bc)i$ +If $z_1 = a+bi$ and $z_2=c+di$, then +$z_1 \times z_2 = (ac-bd)+(ad+bc)i$ ### Conjugates @@ -98,8 +93,7 @@ ${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$ To solve $z^2+a^2=0$ (sum of two squares): -$z^2+a^2=z^2-(ai)^2$ -              $=(z+ai)(z-ai)$ +$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$ ## Polar form -- 2.43.2