Complex Numbers And Quadratic Equations question 662
Question: If $ z=\frac{-2( 1+2i )}{3+i} $ where $ i=\sqrt{-1}, $ then argument $ \theta (-\pi <\theta \le \pi ) $ of z is
Options:
A) $ \frac{3\pi }{4} $
B) $ \frac{\pi }{4} $
C) $ \frac{5\pi }{6} $
D) $ -\frac{3\pi }{4} $
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Answer:
Correct Answer: D
Solution:
First, let’s simplify the complex number z:
$ z = \frac{-2(1+2i)}{3+i} = \frac{-2-4i}{3+i} $
To find the argument, we need to get z in the form a + bi. To do this, we multiply numerator and denominator by the complex conjugate of the denominator:
$ z = \frac{(-2-4i)(3-i)}{(3+i)(3-i)} = \frac{-6+2i-12i+4i^2}{9+1} = \frac{-10-10i}{10} = -1-i $
Now we have z in the form a + bi, where a = -1 and b = -1
The argument θ of a complex number a + bi is given by arctan(b/a), but we need to be careful about which quadrant the number is in.
Since both a and b are negative, z is in the third quadrant.
In the third quadrant, we need to add π to arctan(b/a):
$ θ = arctan(\frac{-1}{-1}) + π = arctan(1) + π = \frac{π}{4} + π = \frac{5π}{4} $
However, the question specifies that -π < θ ≤ π, so we need to subtract 2π:
$ \frac{5π}{4} - 2π = -\frac{3π}{4} $
Therefore, the argument θ of z is -3π/4.
The correct answer is D) $ -\frac{3π}{4} $
BETA
