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Question: How do you simplify \(\left( {3 + 8i} \right)\left( { - 2 - i} \right)\) ?...

How do you simplify (3+8i)(−2−i)\left( {3 + 8i} \right)\left( { - 2 - i} \right) ?

Explanation

Solution

The question belongs to the concept of multiplication of complex numbers. In this type of question we multiply the term of the first bracket to the term of the second bracket. After multiplication we do the same arithmetic operations to the given expression as we do in normal multiplication of algebraic expression. As we know that the complex numbers are the combination of the real numbers and imaginary numbers. We know that when a real number is squared then it gives the only positive value, on the other hand if an imaginary number is squared then it gives the negative value. In a complex number an imaginary number is represented in the multiplication of (i)\left( i \right) with the real number. If a complex number is written as (a+ib)\left( {a + ib} \right) then aa is the real part and (ib)\left( {ib} \right) is the imaginary part. The real numbers are denoted by the letter RR. we will use the imaginary number chart to simplify the given expression. When we multiply a real number with (i)\left( i \right) then it circles through four very different values.

Complete step by step solution:
Step: 1 the given multiplication of complex numbers are,
(3+8i)(−2−i)\left( {3 + 8i} \right)\left( { - 2 - i} \right)
Multiply the term of the first bracket to the term of the second bracket.
⇒(3+8i)(−2−i)=(3×−2)−3×i+(8i×−2)−(8i×i)\Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = \left( {3 \times - 2} \right) - 3 \times i + \left( {8i \times - 2} \right) - \left( {8i \times i} \right)
Now multiply the term inside the brackets.
⇒(3+8i)(−2−i)=(3×−2)−3×i+(8i×−2)−(8i×i) ⇒(3+8i)(−2−i)=−6−3i−16i−8i2  \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = \left( {3 \times - 2} \right) - 3 \times i + \left( {8i \times - 2} \right) - \left( {8i \times i} \right) \\\ \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = - 6 - 3i - 16i - 8{i^2} \\\
Step: 2 now substitute the value of i2=1{i^2} = 1 in the given expression.
⇒(3+8i)(−2−i)=−6−3i−16i−8i2 ⇒(3+8i)(−2−i)=−6−3i−16i+8  \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = - 6 - 3i - 16i - 8{i^2} \\\ \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = - 6 - 3i - 16i + 8 \\\
Add the real part to the real part and imaginary part to the imaginary part in the given expression.
⇒(3+8i)(−2−i)=−6−3i−16i+8 ⇒(3+8i)(−2−i)=2−19i  \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = - 6 - 3i - 16i + 8 \\\ \Rightarrow \left( {3 + 8i} \right)\left( { - 2 - i} \right) = 2 - 19i \\\

Therefore the simple form of the given question is 2−19i2 - 19i.

Note:
Add the imaginary part of the expression to the imaginary part and real part of the expression to the real part. Students are advised to remember the value of (i2=−1)\left( {{i^2} = - 1} \right) and use it in the multiplication of complex numbers.