How do you simplify \[(3 + 4i)(3 + 4i)\]
Answer
602.7k+ views
Hint: Here in this question, we have to find the product of 2 complex numbers. The complex number is one form of number. So let we multiply 2 complex numbers which are different from one another and then we use the arithmetic operation that is multiplication and then we simplify.
Complete step by step solution:
A complex number is a combination of the real part and the imaginary part. The imaginary number is represented by “i”. Usually, the complex number is defined as \[a \pm bi\].
Now let us consider the two complex numbers and they are \[3 + 4i\], and \[3 + 4i\]
Now we have to multiply the complex numbers, to multiply the complex numbers we use multiplication. The multiplication is one of the arithmetic operations.
Now we multiply the above 2 complex numbers
\[(3 + 4i).(3 + 4i)\]
Here dot represents the multiplication. First, we multiply the first two terms of the above equation
\[ \Rightarrow 3(3 + 4i) + 4i(3 + 4i)\]
On multiplying we get
\[ \Rightarrow 9 + 12i + 12i + 16{i^2}\]
On simplification we have
\[ \Rightarrow 9 + 14i + 16{i^2}\]
As we know that the value of \[{i^2} = - 1\]. On substituting this we have
\[ \Rightarrow 9 + 14i + 16\left( { - 1} \right)\]
\[ \Rightarrow 9 + 14i - 16\]
On simplification we have
\[ \Rightarrow 14i - 7\].
Therefore, we have \[(3 + 4i)(3 + 4i) = 14i - 7\]
We can also solve the given question by using the standard algebraic formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]. Here a is 3 and b is 4i.
\[(3 + 4i)(3 + 4i) = {(3 + 4i)^2}\]
On substituting the values in the formula we get
\[ = {3^2} + {(4i)^2} + 2(3)(4i)\]
\[ = 9 + 16{i^2} + 24i\]
As we know that the value of \[{i^2} = - 1\]. On substituting this we have
\[ = 9 - 16 + 24i\]
On simplification we have
\[ \Rightarrow 14i - 7\]
Hence we have multiplied.
Thus the required solution is \[14i - 7\].
Note: To multiply we use operation multiplication, multiplication of numbers is different from the multiplication of algebraic expression. In the algebraic expression it involves the both number that is constant and variables. Variables are also multiplied, if the variable is the same then the result will be in the form of exponent.
Complete step by step solution:
A complex number is a combination of the real part and the imaginary part. The imaginary number is represented by “i”. Usually, the complex number is defined as \[a \pm bi\].
Now let us consider the two complex numbers and they are \[3 + 4i\], and \[3 + 4i\]
Now we have to multiply the complex numbers, to multiply the complex numbers we use multiplication. The multiplication is one of the arithmetic operations.
Now we multiply the above 2 complex numbers
\[(3 + 4i).(3 + 4i)\]
Here dot represents the multiplication. First, we multiply the first two terms of the above equation
\[ \Rightarrow 3(3 + 4i) + 4i(3 + 4i)\]
On multiplying we get
\[ \Rightarrow 9 + 12i + 12i + 16{i^2}\]
On simplification we have
\[ \Rightarrow 9 + 14i + 16{i^2}\]
As we know that the value of \[{i^2} = - 1\]. On substituting this we have
\[ \Rightarrow 9 + 14i + 16\left( { - 1} \right)\]
\[ \Rightarrow 9 + 14i - 16\]
On simplification we have
\[ \Rightarrow 14i - 7\].
Therefore, we have \[(3 + 4i)(3 + 4i) = 14i - 7\]
We can also solve the given question by using the standard algebraic formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]. Here a is 3 and b is 4i.
\[(3 + 4i)(3 + 4i) = {(3 + 4i)^2}\]
On substituting the values in the formula we get
\[ = {3^2} + {(4i)^2} + 2(3)(4i)\]
\[ = 9 + 16{i^2} + 24i\]
As we know that the value of \[{i^2} = - 1\]. On substituting this we have
\[ = 9 - 16 + 24i\]
On simplification we have
\[ \Rightarrow 14i - 7\]
Hence we have multiplied.
Thus the required solution is \[14i - 7\].
Note: To multiply we use operation multiplication, multiplication of numbers is different from the multiplication of algebraic expression. In the algebraic expression it involves the both number that is constant and variables. Variables are also multiplied, if the variable is the same then the result will be in the form of exponent.
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