Question

How do you simplify$\left( 2-3i \right)\left( -1+4i \right)$?

Answer 1

This question is from the topic of complex numbers. We will know everything about iota to solve this question. In this question, first we will get to know the values of ${{i}^{1}}$, ${{i}^{2}}$, ${{i}^{3}}$, and ${{i}^{4}}$. After that, we will multiply the terms $\left( 2-3i \right)$ and $\left( -1+4i \right)$ and use the formulas of iota to solve this.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to simplify $\left( 2-3i \right)\left( -1+4i \right)$ or we can say we have to find the product of $\left( 2-3i \right)$ and $\left( -1+4i \right)$.
So, before simplifying the term $\left( 2-3i \right)\left( -1+4i \right)$, let us first know about iota.
The symbol for the term iota is $i$.
The value of iota is square root of negative of one, or we can say $i=\sqrt{-1}$
So, we can write that the square of iota will be negative of one.
Similarly, we can write the cube of iota as negative of iota and the fourth power iota is one.
Hence, in mathematical term, we can say
${{i}^{1}}=i=\sqrt{-1}$;
${{i}^{2}}=-1$;
${{i}^{3}}={{i}^{2}}{{i}^{1}}=-i$; and
${{i}^{4}}={{i}^{2}}{{i}^{2}}=-1\times -1=1$
Now, we will use them in solving this question.
So, let us simplify the term $\left( 2-3i \right)\left( -1+4i \right)$
The multiplication of the terms $\left( 2-3i \right)$and$\left( -1+4i \right)$ can be written as
$\left( 2-3i \right)\left( -1+4i \right)=2\left( -1+4i \right)-3i\left( -1+4i \right)$
After multiplying the terms inside the parenthesis or bracket with the terms outside the brackets, the above equation in the right hand side can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+8i-3i\left( -1 \right)-3i\times 4i$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+8i+3i-12\times i\times i$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+11i-12\times {{i}^{2}}$
As we have seen in the above that, ${{i}^{2}}=-1$, so we can write the above equation as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+11i-12\times \left( -1 \right)$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+11i+12$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=11i+10$

Hence, the simplified value of $\left( 2-3i \right)\left( -1+4i \right)$ is $11i+10$.

Note: We should have a better knowledge in the topic of complex numbers to solve this question. We should know the value and formulas of iota. We can solve this type of question using the foil method. The foil method says that the multiplication of (a+b) and (c+d) is ac+ad+bc+bd where a, b, c, and d are some numbers.
So, using foil method, we can say that
$\left( 2-3i \right)\left( -1+4i \right)=2\times \left( -1 \right)+2\times 4i-3i\times \left( -1 \right)-3i\times 4i$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=-2+8i+3i+12$
The above equation can also be written as
$\Rightarrow \left( 2-3i \right)\left( -1+4i \right)=10+11i$
Hence, we get the same answer from this method too.
So, we can use this method to multiply these types of factors.