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Question: How do you simplify \[\left( 8-3i \right)\left( 6+i \right)\] ?...

How do you simplify (8−3i)(6+i)\left( 8-3i \right)\left( 6+i \right) ?

Explanation

Solution

Problems of complex numbers can be easily solved using the property of complex numbers. We can simplify the given expression treating them as real numbers as the complex numbers are written in algebraic form. While simplifying the given equation we will take the square of imaginary unit ii equals to −1-1 .

Complete step by step answer:
The expression we have is
(8−3i)(6+i)\left( 8-3i \right)\left( 6+i \right)
As we can see that the complex numbers are written in algebraic form we treat the numbers in the given expression as real numbers.
We expand the given expression using the Foil method as shown below
⇒8(6+i)−3i(6+i)\Rightarrow 8\left( 6+i \right)-3i\left( 6+i \right)
We now apply the distributive property in the first term of the above expression and get
⇒8⋅6+8⋅i−3i(6+i)\Rightarrow 8\cdot 6+8\cdot i-3i\left( 6+i \right)
We now apply the distributive property in the last term of the above expression and get
⇒8⋅6+8⋅i−3i⋅6−3i⋅i\Rightarrow 8\cdot 6+8\cdot i-3i\cdot 6-3i\cdot i
Further completing the calculations in the above equations, we get
⇒48+8i−18i−3i1⋅i1\Rightarrow 48+8i-18i-3{{i}^{1}}\cdot {{i}^{1}}
We use the power rule to combine exponents in the last term. According to the power rule aman=am+n{{a}^{m}}{{a}^{n}}={{a}^{m+n}}
Hence, we rewrite the above expression using the power rule as shown below
⇒48+8i−18i−3i1+1\Rightarrow 48+8i-18i-3{{i}^{1+1}}
Adding 11 and 11 we get
⇒48+8i−18i−3i2\Rightarrow 48+8i-18i-3{{i}^{2}}
Now, we rewrite the complex term i2{{i}^{2}} as (−1)\left( -1 \right) in the above expression and get
⇒48+8i−18i−3(−1)\Rightarrow 48+8i-18i-3\left( -1 \right)
Multiplying 33 to (−1)\left( -1 \right) we get
⇒48+8i−18i+3\Rightarrow 48+8i-18i+3
Adding the like terms in the above expression we get
⇒(48+3)+(8i−18i)\Rightarrow \left( 48+3 \right)+\left( 8i-18i \right)
Completing the summation of the above terms we get
⇒51−10i\Rightarrow 51-10i

Therefore, simplifying the expression (8−3i)(6+i)\left( 8-3i \right)\left( 6+i \right) we get 51−10i51-10i.

Note: The complex numbers can be treated as real numbers as they are written in algebraic form but we must keep in mind that the square of imaginary unit ii equals to −1-1 . So, we must do the simplification accordingly. Also, we must be very careful while doing the summations and multiplications so that both the positive and negative signs are taken into account. In this way mistakes can be avoided.