Question

How do you multiply \[(3 - 5i)(3 + 5i)\] ?

Answer 1

Hint: Real numbers and imaginary numbers are the two types of numbers, real numbers are the numbers that can be plotted on a number line while imaginary numbers, as the name suggests, cannot be represented on the number line. Sometimes while solving equations under the square root, we get a negative answer but we know that the square root of a negative number doesn’t exist so we had to think of a way to represent them that’s why we take $ \sqrt { - 1} = i $ where $ i $ is called iota, so any other negative square root can be expressed in terms of iota, for example $ \sqrt { - 5} = \sqrt 5 i $

Complete step-by-step answer:
The expressions like $ 3 + 5i $ are actually equal to $ 3 + \sqrt { - 5} $ and such numbers are called complex numbers as they involve an imaginary number. We can solve the given question with the help of the values of $ i,{i^2}and\,{i^3} $ .
We know that –
 $
  i = \sqrt { - 1} \\
   \Rightarrow {i^2} = - 1 \\
  {i^3} = {i^2}.i \\
   \Rightarrow {i^3} = - i \;
  $
So, solving \[(3 - 5i)(3 + 5i)\] -
\[
  (3 - 5i)(3 + 5i) = {(3)^2} - {(5i)^2} \\
   \Rightarrow (3 - 5i)(3 + 5i) = 9 - 25{i^2} \\
   \Rightarrow (3 - 5i)(3 + 5i) = 9 - 25( - 1) \\
   \Rightarrow (3 - 5i)(3 + 5i) = 34 \;
 \]
Hence, \[(3 - 5i)(3 + 5i) = 34\] .
So, the correct answer is “34”.

Note: For multiplying the terms written in the parentheses like $ (a + b)(c + d) $ we first multiply the first term of the first bracket with the whole second bracket and then multiply the second term with the whole second bracket that is $ (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd $ but there are various identities to make the calculations easier, in this question, we have used an identity according to which the product of two numbers with opposite sign is equal to the difference of the square of the first number and the square of the second number that is $ (a + b)(a - b) = {a^2} - {b^2} $ . There are many such identities that we can use for solving similar questions, for example – $ {(a + b)^2} = {a^2} + {b^2} + 2ab,\,{(a - b)^2} = {a^2} + {b^2} - 2ab,\,etc. $