From e28efc43108787a51ee11a608c27612a6c365035 Mon Sep 17 00:00:00 2001
From: Timo1104 <t.r.vanabswoude@student.tudelft.nl>
Date: Sat, 8 Aug 2020 15:37:17 +0000
Subject: [PATCH] Improved equation alignment and fixed typo

---
 src/1_complex_numbers.md | 12 ++++++------
 1 file changed, 6 insertions(+), 6 deletions(-)

diff --git a/src/1_complex_numbers.md b/src/1_complex_numbers.md
index ca4285b..803ff34 100644
--- a/src/1_complex_numbers.md
+++ b/src/1_complex_numbers.md
@@ -146,7 +146,7 @@ formulas of cosine and sine.
 
 Some useful values of the complex exponential to know by heart are $e^{2{\rm i } \pi} = 1 $, $e^{{\rm i} \pi} = -1 $ and $e^{{\rm i} \pi/2} = {\rm i}$. 
 From the first expression, it also follows that 
-$$e^{{\rm i} (y + 2\pi n)} = e^{{\rm i}\pi} {rm ~ for ~} n \in \mathbb{Z}$$
+$$e^{{\rm i} (y + 2\pi n)} = e^{{\rm i}\pi} {\rm ~ for ~} n \in \mathbb{Z}$$
 As a result, $y$ is only defined up to $2\pi$.
 
 Furthermore, we can define the sine and cosine in terms of complex exponentials:
@@ -158,17 +158,17 @@ Some operations which are common in real analysis are then easily derived for th
 $$z^{n} = \left(r e^{{\rm i} \varphi}\right)^{n} = r^{n} e^{{\rm i} n \varphi}$$
 $$\sqrt[n]{z} = \sqrt[n]{r e^{{\rm i} \varphi} } = \sqrt[n]{r} e^{{\rm i}\varphi/n} $$
 $$\log(z) = log \left(r e^{{\rm i} \varphi}\right) = log(r) + {\rm i} \varphi$$
-$$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2}})$$
+$$z_{1}z_{2} = r_{1} e^{{\rm i} \varphi_{1}} r_{2} e^{{\rm i} \varphi_{2}} = r_{1} r_{2} e^{{\rm i} (\varphi_{1} + \varphi_{2})}$$
 We see that during multiplication, the norm of the new number is the *product* of the norms of the multiplied numbers, and its argument is the *sum* of the arguments of the multiplied numbers. In the complex plane, this looks as follows:
 
 ![image](figures/complex_numbers_12_0.svg)
 
 **Example** Find all solutions solving $z^4 = 1$. 
 Of course, we know that $z = \pm 1$ are two solutions, but which other solutions are possible? We take a systematic approach:
-$$ z = e^{{\rm i} \varphi} \Rightarrow z^4 = e^{4{\rm i} \varphi} = 1 $$
-$$\Leftrightarrow 4 \varphi = n 2 \pi$$
-$$\Leftrightarrow \varphi = 0, \varphi = \frac{\pi}{2}, \varphi = -\frac{\pi}{2}, \varphi = \pi$$
-$$\Leftrightarrow z = 1, z = i, z = -i, z = -1$$
+$$\begin{align} z = e^{{\rm i} \varphi} & \Rightarrow z^4 = e^{4{\rm i} \varphi} = 1 \\
+& \Leftrightarrow 4 \varphi = n 2 \pi \\
+& \Leftrightarrow \varphi = 0, \varphi = \frac{\pi}{2}, \varphi = -\frac{\pi}{2}, \varphi = \pi \\
+& \Leftrightarrow z = 1, z = i, z = -i, z = -1 \end{align}$$
 
 ### Differentiation and integration
 We only consider differentiation and integration over *real* variables. We can then regard the complex ${\rm i}$ as another constant, and use our usual differentiation and integration rules:
-- 
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