# 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.

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.

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:

$A\ =\ \sqrt{x^2+y^2}$

$A\ =\ \sqrt{5^2+{14}^2}$

$A\ =\ \sqrt{221}=14.866$