Rectangular shape - definition, example and explanation (2023)

The rectangular form of the complex numbers is the first form we will encounter when learning about complex numbers. This shape depends on your Cartesian coordinate and you will see why in the next section.

Rectangular forms of complex numbers represent these numbers by emphasizing the real and imaginary parts of the complex number.

Basic operations are much easier when the complex numbers are in rectangular form. It is more intuitive for us to represent complex numbers in rectangular form since we are more familiar with the Cartesian coordinate system.

This article updates our knowledge of:

• The components that make up acomplex number.
• Graph complex numbers on a complex plane.
• Convert complex numbers in rectangular form topolar shape, and vice versa.
• Edit complex numbers in rectangular form.

Be sure to open your notes and review these concepts, as we'll need them as we learn more about complex numbers in rectangular form.

What is a rectangular shape?

The rectangular shape is based on its name: a rectangular coordinate system. This means that they are complex numbers of the form $z = a + bi$, where $a$ is the real part and $bi$ is the imaginary part. These are some examples of complex numbers in rectangular form.

• $-3 + 4i$ : $-3$ represents the real part while $4i$ represents the imaginary part.
• $-6i$: This is an imaginary number that contains only one imaginary part, $-6i$.
• $5$: Since $5$ is a countable number and therefore a real number, $5$ is still a complex number whose imaginary part equals $0$.

Complex numbers of the form $a + bi$ can be represented on a complex plane simply by graphing $(a,b)$, where $a$ is the real axis coordinate and $b$ is the imaginary axis coordinate.

Here is a diagram of how $a + bi$ is represented in a complex plane. As mentioned above, $a$ represents the distance along the real axis and $b$ the distance along the imaginary axis, a similar approach to graphing rectangular coordinates.

The distance to the origin formed by $a + bi$ is equal to $\sqrt{a^2 + b^2}$ or also known as the module or absolute value of the complex number.

How do I convert a rectangular shape?

As mentioned above, the rectangular form is the first form of complex numbers that we are introduced to, but the complex numbers can also be rewritten in their trigonometric or polar form.

These are just a few examples of pairs of complex numbers in their two forms: rectangular and polar. Let's update what we've learned about writing complex numbers in these two forms.

How to convert a rectangular shape to a polar shape?

We have extensively discussed the conversion of complex numbers in rectangular form, $a + bi$, to trigonometric form (also called polar form). Be sure to check your notes or check out the link we've attached in the first section.

This section is a brief summary of what we have learned in the past:

• Find the modulus $r = \sqrt{a^2 + b^2}$ of the complex number.
• Determine the argument, $\theta = \tan^{-1} \dfrac{b}{a}$, and be sure to choose the angle that is in the correct quadrant.
• Using these values, write the complex number in the form $r(\cos \theta + i\sin \theta)$.

How to convert polar shape to rectangular shape?

Converting complex numbers to polar form is much easier since we only need to evaluate the cosine and sine at values ​​other than $\theta$.

• If you get a complex number in the form $r(\cos \theta + i\sin \theta)$, find the values ​​of $\sin \theta$ and $\cos \theta$.
• Distribute $r$ to each of the evaluated values ​​of $\cos\theta$ and $i\sin\theta$.
• Be sure to return the values ​​of the form $a + bi$.

Without worries. We have prepared some examples for you to work on and practice your knowledge of converting complex numbers to polar form.

Summary of the definition and properties of the rectangular shape.

Why not summarize what we've learned so far about complex numbers in rectangle form before we get into the various problems we've prepared?

• The general (or standard) rectangular form of complex numbers is $a + bi$.
• We can convert complex numbers to rectangular form by finding $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1} \dfrac{b}{a}$.
• Remember that when working with complex number equations, the parts of the real numbers and the parts of the imaginary numbers must be the same for the equation to be valid.

We can also do many things when given a complex number in rectangular form, and we have listed a few that we have learned in the past. Don't have your useful notes with you? Don't worry, we've also included some links for you to check out.

• It is easierAddmisubtractcomplex numbers in rectangular form as we combine the parts of real and imaginary numbers.
• yes we can toomultiplymiPull apartcomplex numbers in rectangular form by algebraic manipulation.
• The product of a $a + bi$ and its conjugate, $a – bi$, equals $a^2 + b^2$, which can help simplify the quotient of two complex numbers.

Let's apply everything we've learned in this article and try these sample tasks.

Example 1

Draw the following complex numbers on the complex plane and give the corresponding absolute values.

and. $6 - $6
b. $-4\sqrt{3} – 4i$
C. $-5i$

solution
Since we also need the absolute value of these three complex numbers, why not start with the fact that $|a + bi| = \sqrt{a^2 + b^2}$?

Now that we have the absolute value of the three complex numbers, let's graph the three complex numbers on a complex plane.

• For $6 – 6i$ plot the coordinates $(6, -6)$ or $6$ units down the real axis and six units down along the imaginary axis.
• Similarly, we can graph $-4\sqrt{3} – 4i$ by graphing $(-4\sqrt{3}, -4)$ on the complex plane.
• Since $-5i$ only contains an imaginary part, we plot $-5i$ on the imaginary axis and need to find units of $5$ below the real axis.

Connect each complex number to the origin and label the segment with the corresponding absolute number.

example 2

Evaluate the following operations with the following complex numbers.

a. $(8 – 8i) + (-6 + 12i)$
b. $(-3\square root{3} + 5i) – (4\square root{3} – 6i)$
C. $(-4 + 2i)(-2 – i) + (2- 3i)$

solution
Remember that adding and subtracting complex numbers is similar to adding and subtracting binomials. We combine the terms with the real numbers and the imaginary numbers. In the same way, we combine "same terms".

First let's work on the first element: $(8 – 8i) + (-6 + 12i)$.

$\begin{aligned} (8 – 8i) + (-6 + 12i) &= [8 + (-6)] + (-8 + 12)i\\&= 2 + 6i\end{aligned}$

Take care to carefully space the negative sign when subtracting two complex numbers.

$\begin{alinhado} (-3\sqrt{3} + 5i) – (4\sqrt{3} – 6i) &=-3\sqrt{3} + 5i – 4\sqrt{3} -(-6i )\\&= -3\sqrt{3} + 5i – 4\sqrt{3} + 6i\\&= (-3\sqrt{3} – 4\sqrt{3}) + (5 + 6)i \\&=-7\sqrt{3} + 11i \end{alineado}$

For the third element, first multiply the two complex numbers.

• Apply the FOIL method to disperse the terms.
• Replace $i^2$ with $-1$.
• Combine parts of real and imaginary numbers.

$\begin{alinehado} (-4 +2i)(-2 – i) &= (-4)(-2)+ (-4)(-i) + (2i)(-2) + (2i)( -i)\\&=8 + 4i – 4i- 2i^2\\&=8 + 4i – 4i -2(-1)\\&=8 + 4i – 4i + 2\\&= (8 + 2 ) + (4 -4)i\\&=10 \end{aligned}$

Replace $(-4 + 2i)(-2 – i)$ with your product and simplify the expression further.

$\begin{alinhado} (-4 +2i)(-2 – i) + (2 – 3i) &= 10 + (2 – 3i)\\&= (10 + 2) – 3i\\&= 12 – 3i\end{alinhado}$

If we enumerate the results of the three operations, we have the following:
uma. $(8 – 8i) + (-6 + 12i) = 2 + 6i $
b. $(-3\square root{3} + 5i) – (4\square root{3} – 6i) = -7\square root{3} + 11i $
C. $(-4 + 2i)(-2 – i) + (2- 3i) = 12 – 3i $

Example 3

Convert the following complex numbers from polar to rectangular form.

one. $-4(\cos 90^{\circ} + i\sin 90^{\circ})$
b. $6\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}\right)$
C. $-\sqrt{3} \text{cis} \dfrac{3\pi}{4}$

solution
Evaluate the cosine and sine values ​​inside the parentheses when converting complex numbers to square form. Distribute the modulus to each of the intrinsic values ​​to simplify the expression to one form, $a +bi$.

Starting with $-4(\cos 90^{\circ} + i\sin 90^{\circ})$, $\cos 90^{\circ} = 0$ and $\sin 90^{\circ} = $1. Replace the terms in parentheses with these values ​​and distribute $-4$.

$\begin{aligned} -4(\cos 90^{\circ} + i\sin 90^{\circ}) &= -4(0 + i)\\&=0 – 4i\\&= -4i \end{aligned}$

For the second point, we need to follow a similar process, but this time we are working with angles in the form of $\pi$. Remember that $\cos \dfrac{\pi}{3} = \dfrac{1}{2}$ and $\sin \dfrac{\pi}{3}= \dfrac{\sqrt{3}} {2} $

$\begin{aligned} 6\left(\cos \dfrac{\in}{3} + i\sin \dfrac{\in}{3}\right) &= 6\left( \dfrac{1}{2 } + i\dfrac{\sqrt{3}}{2}\direct)\\&=6 \cdot \dfrac{1}{2} – 6 \cdot in \dfrac{\sqrt{3}}{2} \\&= 3 – 3\sqrt{3}i\end{aligned}$

For the third element, be sure to rewrite $r \text{cis } \theta$ to $r(\cos \theta + i \sin \theta)$ .

$\begin{aligned} -\sqrt{3} \text{cis} \dfrac{3\pi}{4} &= -\sqrt{3}\left(\cos \dfrac{3\pi}{4} + i\sin \dfrac{3\pi}{4}\direct)\\ &= -\sqrt{3}\left( -\dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt {2}}{2}\direct)\\&=-\sqrt{3} \cdot -\dfrac{\sqrt{2}}{2} – \sqrt{3} \cdot in \dfrac{\sqrt{ 2}}{2}\\&= \dfrac{\sqrt{6}}{2}-i\dfrac{\sqrt{3}}{2}\end{aligned}$

Thus we have the following complex numbers in their rectangular forms:

uma $-4(\cos 90^{\circ} + i\sin 90^{\circ}) = -4i$
b. $6\left(\cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}\right) = 3 – 3\sqrt{3}i$
C. $-\sqrt{3} \text{cis} \dfrac{3\pi}{4} = \dfrac{\sqrt{6}}{2}-i\dfrac{\sqrt{3}}{2 }$