Answer
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Hint: To simplify the given complex number, we need to multiply the numerator and denominator by the conjugate of the denominator. Here, the conjugate of 3i is -3i, so we will multiply the numerator and denominator by -3i and simplify. Then we know that the value of ${{i}^{2}}=-1$ Hence we will substitute this in the obtained equation. Now we will simplify the expression further and write it in the form of a + ib.
Complete step by step answer:
We know that, $i=\sqrt{-1}$ , which is known as “iota” is an imaginary number. To get a real number we need to square it to remove the under-root from -1.
$\begin{align}
& \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\
& \Rightarrow {{i}^{2}}=-1 \\
\end{align}$
To simplify the given expression in question we need to make the denominator a real number. For that to happen we need to multiply the numerator and denominator with the conjugate of the complex number “3i”.
To get the conjugate of a complex number, one has to change the sign of the imaginary part and let the sign of the real part remain as it is.
That means, Complex number = Real part + Imaginary part.
So, its Conjugate will be Complex number = Real part – Imaginary part.
That’s why, here we will get the conjugate of “3i” as “-3i”. So, the expression will look like
$\Rightarrow \left( \dfrac{-2-5i}{3i} \right)\times \left( \dfrac{-3i}{-3i} \right)$
or we can write it as,
$\Rightarrow \dfrac{\left( -2-5i \right)\left( -3i \right)}{\left( 3i \right)\left( -3i \right)}$
which will further give us,
$\begin{align}
& \Rightarrow \dfrac{-2\left( -3i \right)-5i\left( -3i \right)}{-9{{i}^{2}}} \\
& \Rightarrow \dfrac{6i+15{{i}^{2}}}{-9{{i}^{2}}} \\
\end{align}$
Now we know that ${{i}^{2}}=-1$ hence we get the expression as
Hence, we can write it as,
$\begin{align}
& \Rightarrow \dfrac{6i-15}{9} \\
& \Rightarrow -\dfrac{15}{9}+\dfrac{6}{9}i \\
& \Rightarrow -\dfrac{5}{3}+\dfrac{2}{3}i \\
\end{align}$
Thus, the answer is $-\dfrac{5}{3}+\dfrac{2}{3}i$ .
Note:
Now note that the conjugate of a complex number a + ib is defined as a – ib. Hence the conjugate of a pure complex number is just negative of the original number and the conjugate of a pure real number is the number itself. Also conjugate of a conjugate is the number itself.
Complete step by step answer:
We know that, $i=\sqrt{-1}$ , which is known as “iota” is an imaginary number. To get a real number we need to square it to remove the under-root from -1.
$\begin{align}
& \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\
& \Rightarrow {{i}^{2}}=-1 \\
\end{align}$
To simplify the given expression in question we need to make the denominator a real number. For that to happen we need to multiply the numerator and denominator with the conjugate of the complex number “3i”.
To get the conjugate of a complex number, one has to change the sign of the imaginary part and let the sign of the real part remain as it is.
That means, Complex number = Real part + Imaginary part.
So, its Conjugate will be Complex number = Real part – Imaginary part.
That’s why, here we will get the conjugate of “3i” as “-3i”. So, the expression will look like
$\Rightarrow \left( \dfrac{-2-5i}{3i} \right)\times \left( \dfrac{-3i}{-3i} \right)$
or we can write it as,
$\Rightarrow \dfrac{\left( -2-5i \right)\left( -3i \right)}{\left( 3i \right)\left( -3i \right)}$
which will further give us,
$\begin{align}
& \Rightarrow \dfrac{-2\left( -3i \right)-5i\left( -3i \right)}{-9{{i}^{2}}} \\
& \Rightarrow \dfrac{6i+15{{i}^{2}}}{-9{{i}^{2}}} \\
\end{align}$
Now we know that ${{i}^{2}}=-1$ hence we get the expression as
Hence, we can write it as,
$\begin{align}
& \Rightarrow \dfrac{6i-15}{9} \\
& \Rightarrow -\dfrac{15}{9}+\dfrac{6}{9}i \\
& \Rightarrow -\dfrac{5}{3}+\dfrac{2}{3}i \\
\end{align}$
Thus, the answer is $-\dfrac{5}{3}+\dfrac{2}{3}i$ .
Note:
Now note that the conjugate of a complex number a + ib is defined as a – ib. Hence the conjugate of a pure complex number is just negative of the original number and the conjugate of a pure real number is the number itself. Also conjugate of a conjugate is the number itself.
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