Fp1. Complex numbers.

Given that (z+2i)/ (z- lambda i)= i, where lambda is positive, real constant.
Show that z= (lambda/2 +1)+ i( lambda/2 -1)
Given also that tan(arg z)= 0.5 calculate the value of lambda and the value of modulus of z squared.
What knowledge is the first part of the question touching on? And the tan(Arg)- does this mean the argument of the tan of the argument?
Thanks.

Reply 1

ghostwalker

Original post by MathMeister

Rearrangement of formula, then making the denominators real. I.e. multiply top & bottom by conjugate.

And the tan(Arg)- does this mean the argument of the tan of the argument?
Thanks.

tan(arg(z)) means the tan of the arg of z.

If z=x+iy, then tan(arg(z)) = y/x

Reply 2

MathMeister

Original post by ghostwalker

...

I still don't see how. I've multiplied by the complex conjugate and turned the denominator real but this didn't work. I also switched the denominator to i and multiplied by i but this didn't work either :/
Help plz

Reply 3

ghostwalker

Original post by MathMeister

Start by rearranging to make z the subject.

Where's your working?

Reply 4

MathMeister

Original post by ghostwalker

Start by rearranging to make z the subject.
Where's your working?

I cannot upload.
I got i= (z^2 +z lambda i+ 2zi-2lambda)/ (z^2+lambda^2)

Reply 5

ghostwalker

Original post by MathMeister

I cannot upload.

What's that got to do with anything?

I got i= (z^2 +z lambda i+ 2zi-2lambda)/ (z^2+lambda^2)

Don't know how you got that.

From the beginning, multiply through by z- lambda i to start with.

Reply 6

MathMeister

Original post by ghostwalker

What's that got to do with anything?

Don't know how you got that.

From the beginning, multiply through by z- lambda i to start with.

I did that and got that z= i(lambda-1)
I factored z (because I multiplied) and saw what equalled 0 and then I rearranged to get z and factored the i.

(edited 9 years ago)

Reply 7

ghostwalker

Original post by MathMeister

I did that and got that z= i(lambda-1)
I factored z (because I multiplied) and saw what equalled 0 and then I rearranged to get z and factored the i.

Without the working I can't tell where you're going wrong. And do try LaTex.

Reply 8

MathMeister

Original post by ghostwalker

Without the working I can't tell where you're going wrong. And do try LaTex.

I have tried. I know it's slightly crossing the forum rules but can you please post a solution?

Reply 9

davros

Original post by MathMeister

I did that and got that z= i(lambda-1)
I factored z (because I multiplied) and saw what equalled 0 and then I rearranged to get z and factored the i.

If you forget for the moment that i is the square root of -1 then the first part is just GCSE rearrangement.

Using k instead of lambda so I don't have to Latex ( :smile:

), you basically have:

(z + 2i)/(z - ki) = i

Multiply both sides by z-ki to get

z+ 2i = i(z - ki) = iz + k (using i^2 = -1)

Now put all the z terms on one side:
z - iz = k -2i
or
z(1 - i) = k - 2i

At this point I would multiply both sides by 1+i because I can see it will make the LHS quite neat! Can you finish it from here?