Question

How do you simplify $\dfrac{{ - 3 + 2i}}{{2 - 5i}}?$

Answer 1

Hint: As we know that , the standard complex number is in the form of $x + iy$ , therefore we need to convert $\dfrac{{ - 3 + 2i}}{{2 - 5i}}$ , in the standard form of a complex number. We will do that with the help of a rationalization method. In other words, we can say that we rationalize a fraction by multiplying and dividing the fraction with the conjugate of the denominator. We used the formula of conjugate of a complex number to solve this problem. The formula is if a complex number $z = a + ib$ , then the conjugate of this complex number is $\overline z = a - ib$ .

Complete step by step answer:
First of all in order to convert $\dfrac{{ - 3 + 2i}}{{2 - 5i}}$ , in the standard form of a complex number, we need to rationalize $\dfrac{{ - 3 + 2i}}{{2 - 5i}}$ .
Now, in order to rationalize the above fraction, let us first find the conjugate of the denominator. We find the conjugate of a complex number by changing the sign of the imaginary part.
So we get, $\overline {2 - 5i} = 2 + 5i$ .
Now, first let us take $z = \dfrac{{ - 3 + 2i}}{{2 - 5i}}$
On rationalizing , we get
$ \Rightarrow z = \dfrac{{ - 3 + 2i}}{{2 - 5i}} \times \dfrac{{2 + 5i}}{{2 + 5i}}$
This becomes as
$ \Rightarrow z = \dfrac{{( - 3 + 2i)(2 + 5i)}}{{(2 - 5i)(2 + 5i)}}$
By simplifying the brackets in numerator and denominator, we get
$ \Rightarrow z = \dfrac{{ - 3(2 + 5i) + 2i(2 + 5i)}}{{{{(2)}^2} - {{(5i)}^2}}}$
In the denominator, we use $(a + b)(a - b) = {a^2} - {b^2}$
$ \Rightarrow z = \dfrac{{ - 6 - 15i + 4i + 10{i^2}}}{{4 - 25{i^2}}}$
Here we use ${i^2} = - 1$ and simplify the numerator and denominator using the BODMAS rule
$ \Rightarrow z = \dfrac{{ - 6 - 15i + 4i + 10( - 1)}}{{4 - 25( - 1)}}$
$ \Rightarrow z = \dfrac{{ - 6 - 15i + 4i - 10}}{{4 + 25}}$
$ \Rightarrow z = \dfrac{{ - 16 - 11i}}{{29}}$
Now, in order to convert the above fraction into the standard form of a complex number, we will write the real part and the imaginary part separately. So, this can also be written as
$ \Rightarrow z = - \dfrac{{16}}{{29}} - \dfrac{{11i}}{{29}}$
Hence, we get $z = - \dfrac{{16}}{{29}} - \dfrac{{11i}}{{29}}$ , which is in the standard form of a complex number.
Here, the real part is $\operatorname{Re} (z) = - \dfrac{{16}}{{29}}$ and the imaginary part is $\operatorname{Im} (z) = - \dfrac{{11}}{{29}}$ .

Note:
 $\operatorname{Re} (z)$ stands for the real part of the complex number which is denoted as $z$ and $\operatorname{Im} (z)$ stands for the imaginary part of the complex number which is denoted as $z$ . We often take $z = a + ib$ as the standard form of a complex number, where $\operatorname{Re} (z) = a$ and $\operatorname{Im} (z) = b$ .