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Complex number in rectangular form. What is (1+2i)+(1+3i)?

The purpose of this guide is to solve the given set of complex numbers in rectangular form and find their magnitude, angle, and polar form.

The basic concept behind this article is the Complex Numbers, their Addition or Subtraction, and their Rectangular and Polar forms.

A Complex number can be thought of as a combination of a Real Number and an Imaginary Number, which is usually represented in rectangular form as follows:

\[z=a+ib\]

Where:

$a\ ,\ b\ =\ Real\ Numbers$

$z\ =\ Complex\ Number$

$i\ =\ Iota\ =\ Imaginary\ Number$

Part $a$ of the above equation is called the Real Part, whereas value $ib$ is called the Imaginary Part.

Expert Answer

Given that:

First Complex Number $= 1+2i$

Second Complex Number $= 1+3i$

The sum of two complex numbers $(a+ib)$ and $(c+id)$ in rectangular form is calculated as follows by operating on real and imaginary parts separately:

\[(a+ib)+(c+id)\ =\ (a+c)+i(b+d)\]

By substituting the given complex numbers in above equation, we get:

\[\left(1+2i\right)+\left(1+3i\right)\ =\ \left(1+1\right)+i\left(2+3\right)\]

\[\left(1+2i\right)+\left(1+3i\right)\ =\ 2+5i\]

So:

\[Sum\ of\ Complex\ Numbers\ =\ 2+5i\]

This is the binomial form of the sum of complex numbers represented in $x$ and $y$ coordinates as $x=2$ and $y=5$.

In order to find the magnitude $A$ of the given sum of complex numbers, we will use Pythagoras’s Theorem of Triangles to find the hypotenuse of the Triangular Form of the complex numbers.

\[A^2\ =\ x^2+y^2\]

\[A\ =\ \sqrt{x^2+y^2}\]

By substituting the values of both $x$ and $y$, we get:

\[A\ =\ \sqrt{2^2+5^2}\]

\[A\ =\ \sqrt{4+25}\]

\[A\ =\ \sqrt{29}\]

Hence, the magnitude $A$ of the given sum of complex numbers is $\sqrt{29}$.

The angle of the complex numbers is defined as follows if their real numbers are positive:

\[\tan{\theta\ =\ \frac{y}{x}}\]

By substituting the values of both $x$ and $y$, we get:

\[\tan{\theta\ =\ \frac{5}{2}}\]

\[\theta\ =\ \tan^{-1}{\left(\frac{5}{2}\right)}\]

\[\theta\ =\ 68.2°\]

Euler’s identity can be used to convert Complex Numbers from a rectangular form into a polar form represented as follows:

\[A\angle\theta\ =\ x+iy\]

Where:

\[x\ =\ A\cos\theta \]

\[y\ =\ A\sin\theta \]

Hence:

\[A\angle\theta\ =\ A\cos\theta\ +\ iA\sin\theta \]

\[A\angle\theta\ =\ A(\cos\theta\ +\ i\sin\theta) \]

Substituting the value of $A$ and $\theta$, we get:

\[\sqrt{29}\angle68.2° = 29 [\cos(68.2°) + i \sin(68.2°)]\]

Numerical Result

For the given set of complex numbers in rectangular form $(1+2i)+(1+3i)$

The Magnitude $A$ of the Sum of Complex Numbers is:

\[A\ =\ \sqrt{29}\]

The Angle $\theta$ of Complex Number is:

\[\theta\ =\ 68.2°\]

The Polar Form $A\angle\theta$ of Complex Number is:

\[\sqrt{29}\angle68.2° = 29 [\cos(68.2°) + i \sin(68.2°)]\]

Example

Find the magnitude of the Complex numbers in the rectangular form represented by  $(4+1i)\times(2+3i)$.

Solution

Given that:

First Complex Number $= 4+1i$

Second Complex Number $= 2+3i$

The Multiplication of two complex numbers $(a+ib)$ and $(c+id)$ in rectangular form is calculated as follows:

\[(a+ib)\times(c+id)\ =\ ac+iad+ibc+i^2bd\]

As:

\[i^2={(\sqrt{-1})}^2=-1\]

Hence:

\[(a+ib)\times(c+id)\ =\ ac+i(ad+bc)-bd\]

Now, by substituting the given complex number in above expression for multiplication:

\[(4+1i)\times(2+3i)\ =\ 8+12i+2i+3i^2\]

\[(4+1i)\times(2+3i)\ =\ 8+14i-3\ =\ 5+14i\]

By using Pythagoras’ Theorem:

\[A\ =\ \sqrt{x^2+y^2}\]

\[A\ =\ \sqrt{5^2+{14}^2}\]

\[A\ =\ \sqrt{221}=14.866\]

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