What is the complex conjugate of $\left( 7+2i \right)$ ?
Answer
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Hint: A complex conjugate in mathematics is formed by changing the sign of the imaginary term in binomial. Or, one can say it is the reflection of that complex number about the real axis on the Argand plane. When the iota $\left( i \right)$ of a complex number is replaced by the negative of iota $\left( -i \right)$, we get the conjugate of that complex number that shows the image of the complex number.
Complete step by step answer:
We have been given the complex number in the problem as: $\left( 7+2i \right)$
Now, we will first understand what is the real part of this expression and what is the complex (or, imaginary) part of this expression.
Let us say, we have complex number which is given by: $\left( a+ib \right)$
Here, we can see that the complex number is made up of two terms: $\left( a \right)\text{ and }\left( ib \right)$
The term which contains only a real number, in this case, that is $\left( a \right)$ is termed as the real part of the complex number. Mathematically, this is written as:
$\Rightarrow R\left[ a+ib \right]=a$
And, the term which is represented by multiplying it with iota $\left( i \right)$ is the imaginary part of the complex number. Mathematically, this is written as:
$\Rightarrow I\left[ a+ib \right]=b$
So, in order to get the conjugate of this complex number, we need to reverse the sign of the term containing iota, that is:
Conjugate of $(a+ib)$ is equal to $(a-ib)$.
Applying this principle in our expression, which is equal to $\left( 7+2i \right)$, we have the Conjugate of $(7+2i)$ is equal to:
$\begin{align}
& =\left[ 7+2\left( -i \right) \right] \\
& =\left( 7-2i \right) \\
\end{align}$
Hence, the complex conjugate of $\left( 7+2i \right)$ comes out to be $\left( 7-2i \right)$.
Note: The expression for complex conjugate of number has a standard notation. It is represented by writing a dash over the given complex number. Suppose, we have the complex number $\left( a+ib \right)$, then its complex conjugate is represented by writing $\overline{\left( a-ib \right)}$.
Complete step by step answer:
We have been given the complex number in the problem as: $\left( 7+2i \right)$
Now, we will first understand what is the real part of this expression and what is the complex (or, imaginary) part of this expression.
Let us say, we have complex number which is given by: $\left( a+ib \right)$
Here, we can see that the complex number is made up of two terms: $\left( a \right)\text{ and }\left( ib \right)$
The term which contains only a real number, in this case, that is $\left( a \right)$ is termed as the real part of the complex number. Mathematically, this is written as:
$\Rightarrow R\left[ a+ib \right]=a$
And, the term which is represented by multiplying it with iota $\left( i \right)$ is the imaginary part of the complex number. Mathematically, this is written as:
$\Rightarrow I\left[ a+ib \right]=b$
So, in order to get the conjugate of this complex number, we need to reverse the sign of the term containing iota, that is:
Conjugate of $(a+ib)$ is equal to $(a-ib)$.
Applying this principle in our expression, which is equal to $\left( 7+2i \right)$, we have the Conjugate of $(7+2i)$ is equal to:
$\begin{align}
& =\left[ 7+2\left( -i \right) \right] \\
& =\left( 7-2i \right) \\
\end{align}$
Hence, the complex conjugate of $\left( 7+2i \right)$ comes out to be $\left( 7-2i \right)$.
Note: The expression for complex conjugate of number has a standard notation. It is represented by writing a dash over the given complex number. Suppose, we have the complex number $\left( a+ib \right)$, then its complex conjugate is represented by writing $\overline{\left( a-ib \right)}$.
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