Question

How do you simplify $\left( 9+8i \right)+\left( 8-9i \right)$ ?

Answer 1

We are asked to simplify $\left( 9+8i \right)+\left( 8-9i \right)$
To simplify the given term we will learn how complex numbers are written, how they are arranged, what ‘I’ stand for, what is the various power of iota.
Use ${{i}^{2}}=-1$ to simplify our given problem; we will learn that when we add two complex numbers we add real part to real part and imaginary part of one complex number to another complex number. Using these knowledge we solve our problem.

Complete step by step answer:
We are given $\left( 9+8i \right)+\left( 8-9i \right)$, we are asked to simplify.
Before we simplify, we will learn about complex numbers.
We know the term $a+ib$ is known as a complex number. Where ‘I’ is known as iota, and ‘b’ are any real numbers.
Iota (i) is defined as $i=\sqrt{-1}$
Iota has properly that $i'=i,{{i}^{2}}=-1,{{i}^{3}}=-i\text{ and }{{i}^{4}}=1$
We should know that for any complex number $a+ib$ conjugate of it is given as $a-ib$

We are given $\left( 9+8i \right)+\left( 8-9i \right)$.
For any complex $a+ib$ , ‘a’ is called the real part while ‘b’ is called the imaginary part.
To add any complex we will first judge which part is real and which one imaginary.
While addition of complex we just add real part to real part and imaginary of first complex to the imaginary part of second complex numbers.
For example: if we have $a+ib$ and $c+id$
We have $a+ib$ has real part as ‘a’ and imaginary part as ‘b’ and in $c+id$ , ‘c’ is real part while ‘d’ is imaginary part.
So,
$\left( \text{ }a+ib \right)+\left( c+id \right)=\left( a+c \right)+i\left( b+a \right)$
Now, we are asked to add $\left( 9+8i \right)+\left( 8-9i \right)$
In the first number ‘9’ is the real part while in the other part ‘8’ is the real part so they are added up.
And ‘8’ is the imaginary part in the first number and ‘-9’ is imaginary put in the second number so they are added together.
So,
$\begin{aligned} & \left( 9+8i \right)+\left( 8-9i \right)=9+8i+8-9i \\\ & =9+8+\left( 8-9 \right)i \\\ \end{aligned}$

So, we get –
$\left( 9+8i \right)+\left( 8-9i \right)=17-i$

Note: While simplifying, always remember that in product we multiply each term of the first bracket by each term of the other bracket.
Do not make error like –
$\left( a+b \right)\left( c+d \right)=ac+b$ this is wrong.
Also while simplifying things in addition or subtraction we have to apply the sign that is outside the bracket to all terms of the bracket.
Do not make mistake like –
$\left( a+b \right)-\left( c+d \right)=a+b-c+d$ this is wrong.
It should be like –
$\left( a+b \right)-\left( c+d \right)=a+b-c-d$ .