Maths Olympiad Prep

Library / /104 of 520

Algebra Difficulty 2.8 Junior Find the answer

Find the difference between the real part and the imaginary part of the complex number i(6+i)34i\frac{i(-6+i)}{|3-4i|}. The options are:

Pick one

Solution

First, we simplify the given complex number.
i(6+i)34i=6i+i232+(4)2=16i5=1565i\frac{i(-6+i)}{|3-4i|} = \frac{-6i+i^{2}}{\sqrt{3^{2}+(-4)^{2}}} = \frac{-1-6i}{5} = -\frac{1}{5} - \frac{6}{5}i

The real part of the complex number is 15-\frac{1}{5}, and the imaginary part is 65-\frac{6}{5}. The difference between the real part and the imaginary part is:
(15)(65)=1(-\frac{1}{5}) - (-\frac{6}{5}) = 1

Therefore, the correct answer is B\boxed{B}.

This problem involves the algebraic form of complex numbers, their multiplication and division, and the concept of the modulus of a complex number. It is a basic problem that tests fundamental understanding of complex numbers.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.