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How do you write the complex conjugate of the complex number $2ix-i$?

Answer
VerifiedVerified
444.6k+ views
Hint: The complex conjugate of a complex number is defined as the one which has its real part equal to and the imaginary part equal in magnitude but opposite in magnitude to that of the original complex number. So the complex conjugate is obtained by replacing $i$ by $-i$ in the complex number. Therefore, we must take $i$ common from the given complex number so as to write it as $i\left( 2x-1 \right)$. Then on replacing $i$ by $-i$, we will finally obtain the required complex conjugate of the given complex number.

Complete step by step answer:
Let the complex number given in the above question be equal to the complex variable z so that we can write the below equation
$\Rightarrow z=~2ix-i$
We know that the complex conjugate of a complex number has the real part equal to that of the original one, but has the imaginary part equal in magnitude but opposite in sign to that of the original complex number. So the complex conjugate of a complex number can be obtained by replacing $i$ by $-i$. Therefore, we take $i$ common in the above complex number to write it as
$\Rightarrow z=~i\left( 2x-1 \right)$
Now, we replace $i$ by $-i$ in the above complex number to obtain the complex conjugate as
\[\begin{align}
  & \Rightarrow \overline{z}=-i\left( 2x-1 \right) \\
 & \Rightarrow \overline{z}=i\left( 1-2x \right) \\
\end{align}\]

Hence, we have finally obtained the complex conjugate of the given complex number as \[i\left( 1-2x \right)\].

Note: While obtaining the complex conjugate of the given complex number, which is equal to $2ix-i$, we have assumed x to be a real number. We can also write the given complex number in the standard form of $a+ib$, Since both terms of the given complex number contain $i$, its real part is equal to zero and thus it will be written as $0+i\left( 2x-1 \right)$.