
How do you multiply \[(4 + 5i)(4 - 5i)\]?
Answer
556.2k+ views
Hint: Here this question we have to multiply these two terms. The both terms are in the form of complex numbers. The complex number is a combination of real part and complex part. It can also be determined by using the standard algebraic formula \[(a + b)(a - b) = {a^2} - {b^2}\]. Hence we can determine a solution for the given question.
Complete step-by-step answer:
The numbers which are present in the question are in the form of complex numbers. The complex number is a combination of real part and imaginary part. The complex number is represented as \[(a \pm bi)\], here is a real part and the term involving i, is an imaginary part.
Method 1:
Let multiply these two terms \[(4 + 5i)(4 - 5i)\]
\[ \Rightarrow 4(4 - 5i) + 5i(4 - 5i)\]
On multiplying we get
\[ \Rightarrow 16 - 20i + 20i - 25{i^2}\]
On simplifying we have
\[ \Rightarrow 16 - 25{i^2}\]
In the complex number we know that \[{i^2} = - 1\]
Substituting this value in the above equation we have
\[ \Rightarrow 16 - 25( - 1)\]
On simplifying we have
\[ \Rightarrow 16 + 25\]
Add 16 and 25 we get
\[ \Rightarrow 41\]
Therefore \[(4 + 5i)(4 - 5i) = 41\]
Method 2:
Solving by using the standard algebraic formula \[(a + b)(a - b) = {a^2} - {b^2}\]. Consider the given question \[(4 + 5i)(4 - 5i)\] . The question is in the form of \[(a + b)(a - b)\]. By using the formula we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = {(4)^2} - {(5i)^2}\]
Squaring these numbers we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 - 25{i^2}\]
In the complex number we know that \[{i^2} = - 1\]
Substituting this value in the above equation we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 - 25( - 1)\]
On simplifying we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 + 25\]
Add 16 and 25 we get
\[ \Rightarrow (4 + 5i)(4 - 5i) = 41\]
Therefore \[(4 + 5i)(4 - 5i) = 41\]
Hence we have multiplied the two terms and obtained the solution.
So, the correct answer is “41”.
Note: In mathematics we have different forms of numbers namely, natural numbers, whole numbers, integers, rational numbers, irrational numbers, real numbers, complex numbers. The arithmetic operations like addition, subtraction, multiplication and division are applied to these numbers. The numbers which are in the braces we apply the multiplication operation.
Complete step-by-step answer:
The numbers which are present in the question are in the form of complex numbers. The complex number is a combination of real part and imaginary part. The complex number is represented as \[(a \pm bi)\], here is a real part and the term involving i, is an imaginary part.
Method 1:
Let multiply these two terms \[(4 + 5i)(4 - 5i)\]
\[ \Rightarrow 4(4 - 5i) + 5i(4 - 5i)\]
On multiplying we get
\[ \Rightarrow 16 - 20i + 20i - 25{i^2}\]
On simplifying we have
\[ \Rightarrow 16 - 25{i^2}\]
In the complex number we know that \[{i^2} = - 1\]
Substituting this value in the above equation we have
\[ \Rightarrow 16 - 25( - 1)\]
On simplifying we have
\[ \Rightarrow 16 + 25\]
Add 16 and 25 we get
\[ \Rightarrow 41\]
Therefore \[(4 + 5i)(4 - 5i) = 41\]
Method 2:
Solving by using the standard algebraic formula \[(a + b)(a - b) = {a^2} - {b^2}\]. Consider the given question \[(4 + 5i)(4 - 5i)\] . The question is in the form of \[(a + b)(a - b)\]. By using the formula we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = {(4)^2} - {(5i)^2}\]
Squaring these numbers we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 - 25{i^2}\]
In the complex number we know that \[{i^2} = - 1\]
Substituting this value in the above equation we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 - 25( - 1)\]
On simplifying we have
\[ \Rightarrow (4 + 5i)(4 - 5i) = 16 + 25\]
Add 16 and 25 we get
\[ \Rightarrow (4 + 5i)(4 - 5i) = 41\]
Therefore \[(4 + 5i)(4 - 5i) = 41\]
Hence we have multiplied the two terms and obtained the solution.
So, the correct answer is “41”.
Note: In mathematics we have different forms of numbers namely, natural numbers, whole numbers, integers, rational numbers, irrational numbers, real numbers, complex numbers. The arithmetic operations like addition, subtraction, multiplication and division are applied to these numbers. The numbers which are in the braces we apply the multiplication operation.
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