
How do you divide \[\dfrac{i}{2-3i}\] ?
Answer
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Hint: These types of complex number problems are pretty straight forward and very easy to solve. The \[i\] in complex number is known as “iota”, and it is defined as,
\[i=\sqrt{-1}\] and \[{{i}^{2}}=-1\] . Anything which is written of the form \[a+ib\] represents a complex number, where ‘a’ is the real part and ‘b’ is the imaginary part. The above problem can easily be solved by rationalization of the denominator by multiplying the denominator and numerator with the conjugate of the denominator and then problem reduces to multiplication of two complex numbers, and it is done by using normal algebraic multiplication, the addition part is done by adding only the like terms together, that is, adding only the real parts together and the imaginary parts together.
Complete step by step answer:
Now, starting off with the solution of the problem, we multiply the denominator and numerator with the conjugate of the denominator and we get,
\[\begin{align}
& \dfrac{i}{2-3i} \\
& \Rightarrow \dfrac{i\left( 2+3i \right)}{\left( 2-3i \right)\left( 2+3i \right)} \\
\end{align}\]
Now, multiplying the complex numbers on the numerator and the denominator, we get,
\[\Rightarrow \dfrac{2i+3{{i}^{2}}}{4-9{{i}^{2}}}\]
Now, we replace the value of \[{{i}^{2}}=-1\] , we get,
\[\Rightarrow \dfrac{2i+3\left( -1 \right)}{4-9\left( -1 \right)}\]
Now, taking the negative sign out of the bracket we get,
\[\Rightarrow \dfrac{2i-3}{4+9}\]
Now, evaluating we get,
\[\Rightarrow \dfrac{2i-3}{13}\]
We now try to separate the real and the imaginary part, and on separating them, we can further write the equation as,
\[\begin{align}
& \Rightarrow \dfrac{2}{13}i+\dfrac{-3}{13} \\
& \Rightarrow \dfrac{2}{13}i-\dfrac{3}{13} \\
\end{align}\]
Note: We should be very careful about multiplying the conjugate on the numerator and the denominator. The multiplications and additions of the complex numbers should also be done accordingly. Once the answer is achieved, we must write the complex number separately, of the form real part and imaginary part.
\[i=\sqrt{-1}\] and \[{{i}^{2}}=-1\] . Anything which is written of the form \[a+ib\] represents a complex number, where ‘a’ is the real part and ‘b’ is the imaginary part. The above problem can easily be solved by rationalization of the denominator by multiplying the denominator and numerator with the conjugate of the denominator and then problem reduces to multiplication of two complex numbers, and it is done by using normal algebraic multiplication, the addition part is done by adding only the like terms together, that is, adding only the real parts together and the imaginary parts together.
Complete step by step answer:
Now, starting off with the solution of the problem, we multiply the denominator and numerator with the conjugate of the denominator and we get,
\[\begin{align}
& \dfrac{i}{2-3i} \\
& \Rightarrow \dfrac{i\left( 2+3i \right)}{\left( 2-3i \right)\left( 2+3i \right)} \\
\end{align}\]
Now, multiplying the complex numbers on the numerator and the denominator, we get,
\[\Rightarrow \dfrac{2i+3{{i}^{2}}}{4-9{{i}^{2}}}\]
Now, we replace the value of \[{{i}^{2}}=-1\] , we get,
\[\Rightarrow \dfrac{2i+3\left( -1 \right)}{4-9\left( -1 \right)}\]
Now, taking the negative sign out of the bracket we get,
\[\Rightarrow \dfrac{2i-3}{4+9}\]
Now, evaluating we get,
\[\Rightarrow \dfrac{2i-3}{13}\]
We now try to separate the real and the imaginary part, and on separating them, we can further write the equation as,
\[\begin{align}
& \Rightarrow \dfrac{2}{13}i+\dfrac{-3}{13} \\
& \Rightarrow \dfrac{2}{13}i-\dfrac{3}{13} \\
\end{align}\]
Note: We should be very careful about multiplying the conjugate on the numerator and the denominator. The multiplications and additions of the complex numbers should also be done accordingly. Once the answer is achieved, we must write the complex number separately, of the form real part and imaginary part.
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