How do you simplify $\dfrac{-2-5i}{3i}$ ?
Answer
Verified
444k+ views
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.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE
The area of a 6m wide road outside a garden in all class 10 maths CBSE
What is the electric flux through a cube of side 1 class 10 physics CBSE
If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE
The radius and height of a cylinder are in the ratio class 10 maths CBSE
An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
Write an application to the principal requesting five class 10 english CBSE