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Question: How do you simplify \(\dfrac{{ - 3 + 2i}}{{2 - 5i}}?\)...

How do you simplify −3+2i2−5i?\dfrac{{ - 3 + 2i}}{{2 - 5i}}?

Explanation

Solution

As we know that , the standard complex number is in the form of x+iyx + iy , therefore we need to convert −3+2i2−5i\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+ibz = a + ib , then the conjugate of this complex number is z‾=a−ib\overline z = a - ib .

Complete step by step answer:
First of all in order to convert −3+2i2−5i\dfrac{{ - 3 + 2i}}{{2 - 5i}} , in the standard form of a complex number, we need to rationalize −3+2i2−5i\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, 2−5i‾=2+5i\overline {2 - 5i} = 2 + 5i .
Now, first let us take z=−3+2i2−5iz = \dfrac{{ - 3 + 2i}}{{2 - 5i}}
On rationalizing , we get
⇒z=−3+2i2−5i×2+5i2+5i\Rightarrow z = \dfrac{{ - 3 + 2i}}{{2 - 5i}} \times \dfrac{{2 + 5i}}{{2 + 5i}}
This becomes as
⇒z=(−3+2i)(2+5i)(2−5i)(2+5i)\Rightarrow z = \dfrac{{( - 3 + 2i)(2 + 5i)}}{{(2 - 5i)(2 + 5i)}}
By simplifying the brackets in numerator and denominator, we get
⇒z=−3(2+5i)+2i(2+5i)(2)2−(5i)2\Rightarrow z = \dfrac{{ - 3(2 + 5i) + 2i(2 + 5i)}}{{{{(2)}^2} - {{(5i)}^2}}}
In the denominator, we use (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}
⇒z=−6−15i+4i+10i24−25i2\Rightarrow z = \dfrac{{ - 6 - 15i + 4i + 10{i^2}}}{{4 - 25{i^2}}}
Here we use i2=−1{i^2} = - 1 and simplify the numerator and denominator using the BODMAS rule
⇒z=−6−15i+4i+10(−1)4−25(−1)\Rightarrow z = \dfrac{{ - 6 - 15i + 4i + 10( - 1)}}{{4 - 25( - 1)}}
⇒z=−6−15i+4i−104+25\Rightarrow z = \dfrac{{ - 6 - 15i + 4i - 10}}{{4 + 25}}
⇒z=−16−11i29\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
⇒z=−1629−11i29\Rightarrow z = - \dfrac{{16}}{{29}} - \dfrac{{11i}}{{29}}
Hence, we get z=−1629−11i29z = - \dfrac{{16}}{{29}} - \dfrac{{11i}}{{29}} , which is in the standard form of a complex number.
Here, the real part is Re⁡(z)=−1629\operatorname{Re} (z) = - \dfrac{{16}}{{29}} and the imaginary part is Im⁡(z)=−1129\operatorname{Im} (z) = - \dfrac{{11}}{{29}} .

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