Question

How do you simplify the $4i\left( 6-i \right)$ and write the complex number in standard form.

Answer 1

Hint: Now first we will use distributive property to expand the bracket. Now we know that $i=\sqrt{-1}$ hence we have ${{i}^{2}}=-1$ Now using this we will simplify the expression and write it in the form $a+ib$ . Hence we have the complex number in general form.

Complete step by step solution:
Now let us first understand the concept of complex numbers.
Now we are well familiar with real numbers.
Real numbers consist of rational numbers and irrational numbers.
Now we know that a square of a number is always positive in real.
Hence in real numbers the square root of negative numbers is not defined. Now to define the square root of a negative number we assign an imaginary number.
The imaginary number is denoted by I and the value is given by $i=\sqrt{-1}$ .
Now complex numbers are numbers of the form $a+ib$ where a and b are real and $i=\sqrt{-1}$
Now consider the given expression $4i\left( 6-i \right)$
Now we know according to distributive property that $a\left( b-c \right)=ab-ac$
Hence using this property we get,
$\begin{align}
  & \Rightarrow 4i\left( 6 \right)-\left( 4i \right)\left( i \right) \\
 & \Rightarrow 24i-4{{i}^{2}} \\
\end{align}$
Now note that $i=\sqrt{-1}\Rightarrow {{i}^{2}}=-1$ Hence using this in the above expression we get,
$\begin{align}
  & \Rightarrow 24i-4\left( -1 \right) \\
 & \Rightarrow 24i+4 \\
 & \Rightarrow 4+24i \\
\end{align}$
Hence we have the number in the form of a + ib where a = 4 and b = 24.
Hence the given expression can be written in general form of complex number as $4+24i$

Note:
Now note that a purely real number and a pure imaginary numbers are also complex
number. Let us say we have real number 3. Then it can be written as $3+0i$ similarly 5i can be written as $0+5i$ .