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# Complex Number Question for Edexcel FP2 is driving me mad watch

1. The point P resresents the complex number z, where
|z+i|=sqrt(2)|z+(i/2)|

The transformation T from the z-plane to w-plane is given by
w=[z+isqrt(2)]/[izsqrt(2)+1]

Express as a locus in the w-plane the image of P under the transformation T.

I had shown in a previous part of the question that |z|=1/sqrt(2)

|x+i(y+1)|=sqrt(2)|x+i(y+0.5)|
x2+(y+1)2=2[x2+(y+0.5)2]
x2+y2+2y+1=2x2+2y2+2y+0.5
x2+y2=0.5
sqrt(x2+y2)=|x+iy|=|z|=1/sqrt(2)
***
w=[z+isqrt(2)]/[izsqrt(2)+1]

w=[z+isqrt(2)]/[izsqrt(2)+1]
wizsqrt(2)+w=z+isqrt(2)
z[wisqrt(2)-1]=isqrt(2)-w
z=[isqrt(2)-w]/[wisqrt(2)-1]

1/sqrt(2)=|isqrt(2)-w|/|wisqrt(2)-1|
|wisqrt(2)-1|=sqrt(2)|isqrt(2)-w|
Not sure how to simplify this, and anyway, the answer page says |w-isqrt(2)|=|w+i/sqrt(2)|
2. (Original post by Nuclear Ghost)
The point P resresents the complex number z, where
|z+i|=sqrt(2)|z+(i/2)|

The transformation T from the z-plane to w-plane is given by
w=[z+isqrt(2)]/[izsqrt(2)+1]

Express as a locus in the w-plane the image of P under the transformation T.

I had shown in a previous part of the question that |z|=1/sqrt(2)

|x+i(y+1)|=sqrt(2)|x+i(y+0.5)|
x2+(y+1)2=2[x2+(y+0.5)2]
x2+y2+2y+1=2x2+2y2+2y+0.5
x2+y2=0.5
sqrt(x2+y2)=|x+iy|=|z|=1/sqrt(2)
***
w=[z+isqrt(2)]/[izsqrt(2)+1]

w=[z+isqrt(2)]/[izsqrt(2)+1]
wizsqrt(2)+w=z+isqrt(2)
z[wisqrt(2)-1]=isqrt(2)-w
z=[isqrt(2)-w]/[wisqrt(2)-1]

1/sqrt(2)=|isqrt(2)-w|/|wisqrt(2)-1|
|wisqrt(2)-1|=sqrt(2)|isqrt(2)-w|
Not sure how to simplify this, and anyway, the answer page says |w-isqrt(2)|=|w+i/sqrt(2)|

I find it hard to follow

could you post a picture or tell us where the question comes from please.
3. Your solution is equivalent to the answer page, IMHO. The answer page has a non-rationalised denominator so might not be preferable.

You can use the following rules to convert between them:
|a|.|b| = |ab|
|-z| = |z|

Quick explanation here: from your answer, divide both sides by sqrt(2), and bring that into the left hand modulus. Multiply the left hand side by |-i| (allowed since |-i| = 1). Then 'negate' the right hand modulus - multiplying by |-1|.

The advantage of the mark scheme answer is that you can now express the locus as a perpendicular bisector of two points, if I'm right. Good luck!
4. rearrange for z
mod both sides
LHS: |z|= 1/√2
RHS: write w = u+iv
rearrange to get Cartesian
5. (Original post by ronalddotgl)
Your solution is equivalent to the answer page, IMHO. The answer page has a non-rationalised denominator so might not be preferable.

You can use the following rules to convert between them:
|a|.|b| = |ab|
|-z| = |z|

Quick explanation here: from your answer, divide both sides by sqrt(2), and bring that into the left hand modulus. Multiply the left hand side by |-i| (allowed since |-i| = 1). Then 'negate' the right hand modulus.

The advantage of the mark scheme answer is that you can now express the locus as a perpendicular bisector of two points, if I'm right. Good luck!
Seems to work. Thanks.

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Updated: January 13, 2015
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