Q:if $a$ is real, $b=p+iq,z=x+iy$ and $\bar b=p-iq,\bar z=x-iy$ then show that the equation $b(z+\bar z)+\bar b(\bar z-z)+a=0$ represents two straight lines in the complex plane.
My approach:Simplify the equation i get,\begin{align}b(z+\bar z)+\bar b(\bar z-z)+a & =0 \\
(p+iq)(x+iy+x-iy)+(p-iq)(x-iy-x-iy)+a & =0 \\
(2xp-2yq+a)+i(2xq-2py) & =0
\end{align}
But now i get stuck because i don't know how to show that it represent two straight lines. Any hints or solution will be appreciated.
Thanks in advance.
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emonHR
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2Equate the real & the imaginary parts – lab bhattacharjee Oct 24 '18 at 12:04
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3Does this answer your question? Straight Line Equation in Complex Plane – tryst with freedom Aug 22 '21 at 13:47
2 Answers
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You have yielded :
$$(2xp-2yq+a)+i(2xq-2py) =0$$
Now, for a complex expression $x_1 + y_1 i$ to be equal to zero, both imaginary and real parts must be equal to zero. Thus :
$$2xp-2yq+a = 0 \implies y = \frac{p}{q}x + \frac{a}{2q}$$
$$2xq - 2py = 0 \implies y = \frac{q}{p}x$$
which are both equations of straight lines, as $p,q,a$ are constants $ \in \mathbb R$.
Rebellos
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Hint:
- If $A$ and $B$ are real, $A+Bi=0$ if and only if $A=B=0$.
- $2xp-2yq+a$ and $2xq-2py$ are real.
5xum
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