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<div class="book-prose">
<p>
<strong>Complex numbers</strong> extend the real number system so equations like
\(x^2+1=0\) can be solved. In high school, you usually write a complex number in
<strong>Cartesian form</strong>:
\[
z = a + bi,
\]
where \(a,b\in\mathbb{R}\) and \(i^2=-1\).
</p>
<h2 id="imaginary-unit" class="book-prose__heading">The imaginary unit \(i\)</h2>
<div class="book-callout book-callout--formula" role="note">
<p class="book-callout__label">Definition</p>
<p class="book-formula">\[ i^2 = -1 \]</p>
</div>
<p>
Useful cycle:
\[
i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,
\]
and then the pattern repeats every 4 powers.
</p>
<h2 id="cartesian-form" class="book-prose__heading">Real and imaginary parts</h2>
<p>
For \(z=a+bi\):
</p>
<ul>
<li>\(\Re(z)=a\) (real part)</li>
<li>\(\Im(z)=b\) (imaginary coefficient)</li>
</ul>
<p><strong>Example.</strong> If \(z=3-4i\), then \(\Re(z)=3\), \(\Im(z)=-4\).</p>
<h2 id="operations" class="book-prose__heading">Basic operations</h2>
<p>
Add/subtract by combining like terms:
</p>
<p class="book-formula">
\[
(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i.
\]
</p>
<p>
Multiply using expansion and \(i^2=-1\):
</p>
<p class="book-formula">
\[
(a+bi)(c+di)=ac+adi+bci+bd\,i^2=(ac-bd)+(ad+bc)i.
\]
</p>
<p><strong>Example.</strong> \((2+3i)(1-4i)=2-8i+3i-12i^2=14-5i.\)</p>
<h2 id="conjugate" class="book-prose__heading">Conjugate and modulus</h2>
<p>
The conjugate of \(z=a+bi\) is
\[
\overline{z}=a-bi.
\]
Multiplying by conjugates removes imaginary parts:
\[
z\overline{z}=(a+bi)(a-bi)=a^2+b^2.
\]
</p>
<p>
The modulus (distance from origin in Argand plane) is:
\[
|z|=\sqrt{a^2+b^2}.
\]
</p>
<p><strong>Example.</strong> For \(z=3-4i\), \(\overline{z}=3+4i\), \(|z|=5\).</p>
<h2 id="division" class="book-prose__heading">Division in Cartesian form</h2>
<p>
To divide by \(c+di\), multiply top and bottom by \(c-di\):
</p>
<p class="book-formula">
\[
\frac{a+bi}{c+di}
=\frac{(a+bi)(c-di)}{(c+di)(c-di)}
=\frac{(a+bi)(c-di)}{c^2+d^2}.
\]
</p>
<p><strong>Example.</strong> \(\dfrac{1+2i}{3-i}\):</p>
<p>
\[
\frac{1+2i}{3-i}\cdot\frac{3+i}{3+i}
=\frac{(1+2i)(3+i)}{10}
=\frac{1+7i}{10}
=\frac{1}{10}+\frac{7}{10}i.
\]
</p>
<h2 id="argand-plane" class="book-prose__heading">Argand plane (graph view)</h2>
<p>
Plot \(z=a+bi\) as the point \((a,b)\): horizontal axis is real part, vertical axis is
imaginary part. This lets you interpret \(|z|\) as geometric distance.
</p>
<h2 id="solving-quadratics" class="book-prose__heading">Why this matters: solving quadratics</h2>
<p>
If \(x^2+4=0\), then \(x^2=-4\), so
\[
x=\pm 2i.
\]
Complex numbers ensure every quadratic has two roots (counting multiplicity), and later this
extends to all polynomial equations.
</p>
</div>