Question

How do you simplify\[\dfrac{-2-5i}{3i}\]?

Answer 1

Hint: In order to solve the given expression that contains an imaginary number iota represented by ‘\[i\]’. First we need to write the denominator into a + bi form. Then we multiply numerator and denominator by the conjugate of the denominator to balance the given expression. When we multiply the conjugate to the denominator, it will obtain a real number in the denominator and we will simplify the expression further.

Complete step by step solution:
We have given that,
\[\Rightarrow \dfrac{-2-5i}{3i}\]
Writing the denominator in the given expression into standard form of complex number, we get
Standard form of the complex number is \[a+bi\].
We get,
\[\Rightarrow \dfrac{-2-5i}{0+3i}\]
Multiply the numerator and denominator by the complex conjugate of denominator, we get
Complex conjugate of denominator =\[0-3i\].
\[\Rightarrow \dfrac{-2-5i}{0+3i}\times \dfrac{0-3i}{0-3i}\]
Expanding the numerator and denominator, we get
\[\Rightarrow \dfrac{\left( -2\times 0 \right)+\left( -2\times \left( -3i \right) \right)+\left( -5i\times 0 \right)+\left( -5i\times \left( -3i \right) \right)}{\left( 0\times 0 \right)+\left( 0\times \left( -3i \right) \right)+\left( 3i\times 0 \right)+\left( 3i\times \left( -3i \right) \right)}\]
Simplifying the above by multiplying the numbers in the brackets, we get
\[\Rightarrow \dfrac{0+6i+0+15{{i}^{2}}}{0+0+0+\left( -9{{i}^{2}} \right)}\]
Simplifying the numerator and denominator all adding all the terms, we get
\[\Rightarrow \dfrac{6i+15{{i}^{2}}}{-9{{i}^{2}}}\]
As, we know that
\[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\]
Substituting the value of \[{{i}^{2}}=-1\] in the above given fraction, we get
\[\Rightarrow \dfrac{6i+15\times \left( -1 \right)}{-9\times \left( -1 \right)}\]
Simplifying the above, we get
\[\Rightarrow \dfrac{6i-15}{9}\]
Rewrite the above as,
\[\Rightarrow \dfrac{-15+6i}{9}\]
Expanding the above, we get
\[\Rightarrow -\dfrac{15}{9}+\dfrac{6i}{9}\]
Converting the fractions into simplest form, we get
\[\Rightarrow -\dfrac{5}{3}+\dfrac{2}{3}i\]
Therefore,
\[\Rightarrow \dfrac{-2-5i}{3i}=-\dfrac{5}{3}+\dfrac{2}{3}i\]
Hence, it is the required solution.

Note: Students should always remember that when these types of questions are asked in which we have to simplify the fraction that contains an imaginary expression, it will be suitable to multiply the numerator and denominator both by conjugating the denominator. Always write the answer in the standard form of the complex number i.e. \[a+bi\].