Complex Numbers: Equation involving the argument operator.

AI Thread Summary
The discussion revolves around solving an equation involving the argument operator for complex numbers. The user successfully demonstrates that the left-hand side (LHS) and right-hand side (RHS) of the equation are equal, leading to the conclusion that y = (5/3)x - (1/3). A shortcut method is proposed, utilizing the gradient of the line formed by points z1 and z2, which confirms the same relationship. Both methods are validated as correct, suggesting that practice will enhance efficiency in using either approach. The conversation emphasizes understanding complex number arguments and finding alternative solutions.
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Homework Statement


Question:
Sr5oxE3.png


Homework Equations


Any relevant to complex numbers.

The Attempt at a Solution



Given,

Arg(\frac{z}{w})= Arg(z)-Arg(w)

z=x+yi
z1 = -1-2i
z2 = 2+3i

Arg(z-z1)=Arg(z2-z1)

LHS:
Arg(x+yi+1+2i)
Arg((x+1) + i(y+2))

tan(\theta)=\frac{y+2}{x+1}

RHS
Arg(2+3i+1+2i)
Arg(3+5i)

tan(\theta)=\frac{5}{3}

RHS = LHS

\frac{y+2}{x+1} = \frac{5}{3}

Simplified
y = \frac{5}{3}x -\frac{1}{3}

Is this right?
 
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Yes it is correct.
 
Sunil Simha said:
Yes it is correct.

Cool beans,

Is there a simple way to solve these though?

A shortcut method?Wait, I found it...

It's the line created from the points z1= -1-2i and z2 = 2+3i

Gradient = \frac{3 - (-2)}{2 - (-1)} = \frac{5}{3}

y = \frac{5}{3}x + b

(-2) = \frac{5}{3}(-1) + b

-6 = -5 + 3b
-1 = 3b
b = -\frac{1}{3}
 
Last edited:
Yes. That's how I solved it. Your original method is still fine though. If you have loads of practice,, you'll be able to use both the methods equally efficiently.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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