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I have a mapping $$w=(\sqrt 3+i)z-1+\sqrt3 i$$ and I am trying to find the image of $$y=0$$ under this mapping and I have reached a solution but it is wrong.

This is my working.

I did $$u+iv=(\sqrt 3 +i)(x+iy)-1+\sqrt 3 i$$ Expanding this I get $$(u+iv)=\sqrt 3 x-1-y+i(x+y \sqrt 3+\sqrt 3)$$ So I let $$u=\sqrt 3 x-1-y$$ $$v=x+y \sqrt 3+ \sqrt 3$$

I now let $y=0$ and I rearrange the expression above for $u=...$ This gives me $$x=v-\sqrt 3$$ I also rearrange the expression for $v=..$ for y and I get $y=\sqrt 3x-1-u$ which gives me $$\sqrt 3x-1-u=0$$

and I substitute the equation for x into this to get $$\sqrt 3 v-4+u=0$$

I then solve for $v$ which gives $$v=\frac{4+u}{\sqrt 3}$$

The right answer is $$v=\frac{\sqrt 3}{3}(u+4)$$ and I am not sure how to get this.

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  • $\begingroup$ Note that the map is linear, so it maps lines to lines. Therefore it suffices, e.g., to compute the image under the map of two points on the original line, say, $0$ and $i$. $\endgroup$ Commented May 31, 2015 at 15:29

2 Answers 2

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Hint: $$ \dfrac{4+u}{\sqrt{3}}=\dfrac{\sqrt{3}(4+u)}{3} $$ so your solution is the same as the answer.

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  • $\begingroup$ out of interest, why do we solve for v and not u? $\endgroup$
    – Al jabra
    Commented May 31, 2015 at 15:45
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For any $a \neq 0$ $$\frac{a}{\sqrt{a}}=\sqrt{a}$$

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  • $\begingroup$ I think you should add the "HINT" word at the beginning of your answer. $\endgroup$
    – Daniel
    Commented May 31, 2015 at 15:49

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