Question

How do you divide $\dfrac{1+9i}{-2+9i}$ ?

Answer 1

Hint: When we divide some number by complex number , we convert the denominator to real number by multiplying the conjugate of the denominator to both denominator and numerator. The conjugate of a +ib is equal to a –ib. The product of a + ib and a – ib is equal to ${{a}^{2}}+{{b}^{2}}$ which is a real number.

Complete step by step answer:
We have evaluated $\dfrac{1+9i}{-2+9i}$ , we can see the denominator is -2 + 9i . We have to make it a real number. Conjugate of -2 + 9i is equal to -2 – 9i .
So we can multiply -2 – 9i to both numerator and denominator to make the denominator real.
So multiplying -2 – 9i with -2 + 9i we get ${{\left( -2 \right)}^{2}}+{{\left( -9 \right)}^{2}}$ which is equal to 85
Now we have to multiply -2 – 9i with 1 + 9i , we know that square of i is equal to -1
So $\left( -2-9i \right)\left( 1+9i \right)=79-27i$
So the value of $\dfrac{1+9i}{-2+9i}$ is equal to $\dfrac{79-27i}{85}$

Note:
We can solve it by another method. We can write both complex numbers in numerator and denominator as a product of mods of complex numbers and e to the power argument of the number. The final result will be a product of the power difference between the argument of numerator and denominator and mod of numerator divided by mod of denominator. In this case the denominator should not be equal to 0.