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FP3 Complex numbers

Hi,

I'm told:

w=2z2iw = \dfrac{2}{z - 2i}

I'm asked to show that:

z=2wiw. | z | = 2 \begin{vmatrix} \dfrac{w-i}{w} \end{vmatrix}.

I did this by saying:

w=2z2iw = \dfrac{2}{z - 2i}

z2i=2w\Rightarrow z - 2i = \dfrac{2}{w}

z=2+2wiwz = \dfrac{2+2wi}{w}

z=2(1+wiw)\Rightarrow z = 2(\dfrac{1+wi}{w})

This is where I'm not too sure I've shown what I'm asked to sufficiently. I say this:

z=2[1i(wi)w]z = 2\begin{bmatrix} \dfrac{-\frac{1}{i}(w-i)}{w} \end{bmatrix}

As 1i=1\begin{vmatrix} -\dfrac{1}{i} \end{vmatrix} = 1

z=2wiw.\Rightarrow \begin{vmatrix}z \end{vmatrix} = 2\begin{vmatrix} \dfrac{w-i}{w} \end{vmatrix}.

Would this be a sufficient "show" answer, or does it seem like I've winged it?

Thank you :smile:
Original post by so it goes
Hi,

I'm told:

w=2z2iw = \dfrac{2}{z - 2i}

I'm asked to show that:

z=2wiw. | z | = 2 \begin{vmatrix} \dfrac{w-i}{w} \end{vmatrix}.

I did this by saying:

w=2z2iw = \dfrac{2}{z - 2i}

z2i=2w\Rightarrow z - 2i = \dfrac{2}{w}

z=2+2wiwz = \dfrac{2+2wi}{w}

z=2(1+wiw)\Rightarrow z = 2(\dfrac{1+wi}{w})

This is where I'm not too sure I've shown what I'm asked to sufficiently. I say this:

z=2[1i(wi)w]z = 2\begin{bmatrix} \dfrac{-\frac{1}{i}(w-i)}{w} \end{bmatrix}

As 1i=1\begin{vmatrix} -\dfrac{1}{i} \end{vmatrix} = 1

z=2wiw.\Rightarrow \begin{vmatrix}z \end{vmatrix} = 2\begin{vmatrix} \dfrac{w-i}{w} \end{vmatrix}.

Would this be a sufficient "show" answer, or does it seem like I've winged it?

Thank you :smile:


I personally think what you've done is acceptable
Original post by so it goes
Hi,


Would this be a sufficient "show" answer, or does it seem like I've winged it?



It's a perfect answer.
Reply 3
Original post by MathsNerd1
I personally think what you've done is acceptable



Original post by FireGarden
It's a perfect answer.


Thank you :biggrin:

Sorry I couldn't give you an up-vote FireGarden but TSR's being silly :smile:
Reply 4
Perfectly fine. If you wanted to keep the chain of equals signs going, you could have just split your modulus sign, something like this (since ab=ab|ab|=|a||b| for all a,bCa,b\in\mathbb{C}):

z=2(1i(wi)w)=2(i(wi)w)=2iwiw=2wiw|z| = \left|2\left(\frac{-\frac{1}{i}(w-i)}{w}\right)\right| = \left|2\left(\frac{i(w-i)}{w}\right)\right| = |2i| \left|\frac{w-i}{w}\right| = 2\left|\frac{w-i}{w}\right|

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