1 The Complex Conjugate

In the complex plane, a point $z$ and its conjugate $\overline z$ are symmetric about the real axis.

2 The Negative of a Complex Number

In the complex plane, the points $z$ and $-z$ are symmetric about the origin.

3 Complex Addition

Addition of complex numbers can be visualized in the complex plane by vector addition.

If $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ then the sum is formed by adding the corresponding components: $$z_1 +z_2= (x_1+x_2) + i(y_1+y_2)$$

This can be expressed in terms of real and imaginary parts as $$\Rep (z_1 + z_2) = \Rep (z_1)+ \Rep (z_2)$$ and $$\Imp (z_1+z_2) = \Imp (z_1)+\Imp (z_2)$$

The above figure suggests the triangle inequality, which is proved at the end of the section: $$|z_1 + z_2| \leq |z_1| + |z_2|$$

The modulus of a difference gives the distance between the complex numbers.

4 Complex Multiplication

Using the polar form, we can prove that complex multiplication corresponds to multiplying the moduli and adding the arguments. If $z_1 = r_1 \cis \theta _1$ and $z_2 = r_2 \cis \theta _2$, then $$z_1 z_2 = r_1 r_2\cis \left (\theta _1+\theta _2\right )$$

This follows from the sum identities for the sine and cosine: $$\sin \left (\theta _1+\theta _2\right ) = \sin \theta _1 \cos \theta _2 + \cos \theta _1 \sin \theta _2$$ and $$\cos \left (\theta _1+\theta _2\right ) = \cos \theta _1 \cos \theta _2 - \sin \theta _1 \sin \theta _2$$

In particular, \begin{align*} \cis \theta _1 \cis \theta _2 & = \left ( \cos \theta _1 +i\sin \theta _1 \right ) \left ( \cos \theta _2 + i \sin \theta _2 \right )\\ &= \left (\cos \theta _1 \cos \theta _2 - \sin \theta _1 \sin \theta _2\right )+i \left (\sin \theta _1\cos \theta _2 + \cos \theta _1 \sin \theta _2 \right )\\ &= \cos \left (\theta _1+\theta _2\right ) + i \sin \left (\theta _1+\theta _2\right ) \\ &= \cis \left (\theta _1+\theta _2\right ) \end{align*}

5 Proofs of the Triangle Inequality

The first proof uses the law of cosines: $$|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1|\cdot |z_2|\cos \theta ,$$ where $\theta$ is the angle opposite the side of length $|z_1 + z_2|$. Since $-1 \leq \cos \theta \leq 1$, we have $$|z_1 + z_2|^2 \leq |z_1|^2 + |z_2|^2 + 2|z_1|\cdot |z_2| = \left (|z_1|^2 + |z_2|^2\right ),$$ and the result follows by taking square roots.

The second proof uses the fact that for any real numbers $a$ and $b$, $$2ab \leq a^2 + b^2$$ Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$. Then $z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$ and so $$|z_1 + z_2|^2 = (x_1 +x_2)^2 + (y_1 +y_2)^2$$ Expanding the right hand side gives $$|z_1 + z_2|^2 = x_1^2 +2x_1x_2 +x_2^2 + y_1^2 + 2y_1y_2 +y_2^2 = |z_1|^2 + |z_2|^2 + 2(x_1x_2 + y_1y_2)$$ The result will follow once we show that $(x_1x_2 + y_1y_2)^2 \leq |z_1|^2 |z_2|^2$. Expanding the left hand side gives $$(x_1x_2 + y_1y_2)^2 = (x_1x_2)^2 + 2(x_1x_2y_1y_2) + (y_1y_2)^2$$ Expanding the right hand side gives $$|z_1|^2 |z_2|^2 = (x_1^2+y_1^2)(x_2^2+y_2^2) = (x_1x_2)^2 + x_1^2y_2^2 + x_2^2y_1^2 + (y_1y_2)^2$$ Finally, $$2(x_1x_2y_1y_2) \leq x_1^2y_2^2 + x_2^2y_1^2$$ using $2ab \leq a^2 + b^2$ with $a = x_1y_2$ and $b=x_2y_1$.

The third and final proof uses the following properties of complex numbers:

• $|z|^2 = z{\overline z}$
• ${\overline {z_1 + z_2}}= {\overline z_1} +{\overline z_2}$
• ${\overline {z_1z_2}} = {\overline z_1} {\overline z_2}$
• ${\overline {\overline z}} = z$
• $\Rep (z) = \frac 12 (z+{\overline z})$
• $\Rep (z) \leq |z|$
• $|{\overline z}| = |z|$
• $|z_1z_2| = |z_1|\cdot |z_2|$

We have \begin{align*} |z_1 + z_2|^2 &= (z_1 + z_2)({\overline{z_1 + z_2}}), \; \mbox{by} \; (1)\\ &= (z_1 + z_2)({\overline z_1} +{\overline z_2}), \; \mbox{by} \; (2)\\ &= z_1{\overline z_1} + z_1{\overline z_2}+{\overline z_1}z_2 + z_2{\overline z_2} \\ &= |z_1|^2 + z_1{\overline z_2} +{\overline{z_1{\overline z_2}}} + |z_2|^2 ,\; \mbox{by} \;(3) \, \mbox{and} \, (4)\\ &= |z_1|^2 + 2\Rep (z_1{\overline z_2}) + |z_2|^2, \; \mbox{by} \; (5) \\ & \leq |z_1|^2 + 2 |z_1{\overline z_2}| + |z_2|^2, \; \mbox{by} \;(6)\\ &= |z_1|^2 + 2 |z_1| \cdot |z_2| + |z_2|^2, \; \mbox{by} \;(7) \, \mbox{and} \, (8)\\ &= |z_1 + z_2|^2 \end{align*}

The result follow by taking square roots.