
How do you find the division of the complex number \[\dfrac{4+i}{8+9i}\]?
Answer
547.8k+ views
Hint: In this problem, we have to divide the given fraction. We know that the given fraction has complex numbers. We should know that to divide the fraction we can multiply the numerator/ denominator by the complex conjugate of the denominator. We can then multiply both to get a simplified form of the given problem.
Complete step-by-step solution:
We know that the given fraction is,
\[\dfrac{4+i}{8+9i}\]
We can now find the complex conjugate of the denominator.
the conjugate of the denominator \[8+9i\] is \[8-9i\].
We can now multiply the complex conjugate in both the numerator and the denominator, we get
\[\Rightarrow \dfrac{\left( 4+i \right)}{\left( 8+9i \right)}\times \dfrac{\left( 8-9i \right)}{\left( 8-9i \right)}\]
We can now multiply each term with every term in both the numerator and the denominator, we get
\[\Rightarrow \dfrac{32-36i+8i-9{{i}^{2}}}{64-72i+72i-81{{i}^{2}}}\]
Now we can simplify the above step by cancelling and adding or subtracting similar terms,
\[\Rightarrow \dfrac{32-28i-9{{i}^{2}}}{64-81{{i}^{2}}}\]
We also know that in complex numbers,
\[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\]
We can substitute the above value in the above step, we get
\[\Rightarrow \dfrac{32-28i-\left( -9 \right)}{64-\left( -81 \right)}\]
Now we can simplify the above step, we get
\[\Rightarrow \dfrac{41-28i}{145}\]
We can see that we have two terms in the numerator with one denominator, we can separate it, we get
\[\Rightarrow \dfrac{41}{145}-\dfrac{28}{145}i\]
Therefore, the answer is \[\dfrac{41}{145}-\dfrac{28}{145}i\].
Note: Students make mistakes While taking conjugate for the complex number. We should always remember that the conjugate of the number is changing the sign to its opposite one in the imaginary part. We should also know some complex formulas such as \[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\] to solve these types of problems.
Complete step-by-step solution:
We know that the given fraction is,
\[\dfrac{4+i}{8+9i}\]
We can now find the complex conjugate of the denominator.
the conjugate of the denominator \[8+9i\] is \[8-9i\].
We can now multiply the complex conjugate in both the numerator and the denominator, we get
\[\Rightarrow \dfrac{\left( 4+i \right)}{\left( 8+9i \right)}\times \dfrac{\left( 8-9i \right)}{\left( 8-9i \right)}\]
We can now multiply each term with every term in both the numerator and the denominator, we get
\[\Rightarrow \dfrac{32-36i+8i-9{{i}^{2}}}{64-72i+72i-81{{i}^{2}}}\]
Now we can simplify the above step by cancelling and adding or subtracting similar terms,
\[\Rightarrow \dfrac{32-28i-9{{i}^{2}}}{64-81{{i}^{2}}}\]
We also know that in complex numbers,
\[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\]
We can substitute the above value in the above step, we get
\[\Rightarrow \dfrac{32-28i-\left( -9 \right)}{64-\left( -81 \right)}\]
Now we can simplify the above step, we get
\[\Rightarrow \dfrac{41-28i}{145}\]
We can see that we have two terms in the numerator with one denominator, we can separate it, we get
\[\Rightarrow \dfrac{41}{145}-\dfrac{28}{145}i\]
Therefore, the answer is \[\dfrac{41}{145}-\dfrac{28}{145}i\].
Note: Students make mistakes While taking conjugate for the complex number. We should always remember that the conjugate of the number is changing the sign to its opposite one in the imaginary part. We should also know some complex formulas such as \[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1\] to solve these types of problems.
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