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Question: How do you simplify \(\dfrac{{2 - 3i}}{{3 + 4i}}\)?...

How do you simplify 2−3i3+4i\dfrac{{2 - 3i}}{{3 + 4i}}?

Explanation

Solution

To solve this complex number, take complex conjugate and proceed with normal simplification. And remember some basic formula in the topic complex number like i2=−1{i^2} = - 1. This will help you to solve this problem.

Complete step by step answer:
Let us consider the given solution,
⇒2−3i3+4i\Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}}
Let us take complex conjugate for the above question, to take complex conjugate for a given number, we have to consider the denominator part and have to change the sign near the imaginary part and multiply and divide in the above question we get,
⇒2−3i3+4i=2−3i3+4i×3−4i3−4i   \Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{2 - 3i}}{{3 + 4i}} \times \dfrac{{3 - 4i}}{{3 - 4i}} \\\ \\\
Now use the formula a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) in the above equation and we get,
⇒2−3i3+4i=6−8i−9i+12i29−12i+12i−16i2\Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{6 - 8i - 9i + 12{i^2}}}{{9 - 12i + 12i - 16{i^2}}}
We can perform operation real part with real part and imaginary part with imaginary part only i.e., we cannot add real number with imaginary number We know that i2=−1{i^2} = - 1, substituting the value we get,
⇒2−3i3+4i=6−17i−129+16=−6−17i25=−625+−1725\Rightarrow \dfrac{{2 - 3i}}{{3 + 4i}} = \dfrac{{6 - 17i - 12}}{{9 + 16}} = \dfrac{{ - 6 - 17i}}{{25}} = \dfrac{{ - 6}}{{25}} + \dfrac{{ - 17}}{{25}}
This is our required solution.

Additional information: Some of the properties of complex numbers are

  1. If we add two complex conjugate numbers, it will give a real number.
  2. And also if we multiply two complex conjugate numbers, it will give a real number.
  3. The complex conjugate for z=a+ibz = a + ib will be zˉ=a−ib\bar z = a - ib.
  4. It obeys both addition and multiplication commutative property.

Note: The complex number is expressed in terms of a+iba + ib, where ii is the imaginary number and the other two variables are the real number. When we come up with the solution x=−1x = \sqrt { - 1} , there is no solution for this in real numbers and here we consider the complex number as x=ix = i. Complex numbers will always have real and imaginary parts.