AmTran_Bus
  • AmTran_Bus
Complex numbers
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AmTran_Bus
  • AmTran_Bus
If Z=x+2iy, then find Re(z*) Where z=x+iy is the formula and x=Re(z) and y=IM(z)
AmTran_Bus
  • AmTran_Bus
Re= real , IM=imag.
AmTran_Bus
  • AmTran_Bus
* is the conjugate

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More answers

ganeshie8
  • ganeshie8
\(z^*\) is obtained by reflecting the complex number over real axis it is called the conjugate of \(z\)
ganeshie8
  • ganeshie8
when you reflect something over real axis, notice that the real component is not changed
AmTran_Bus
  • AmTran_Bus
Here is how it says to do it: Re(z*) =Re(x-2iy) =x. Can you help me understand it that way?
ganeshie8
  • ganeshie8
just enter the part that is not attached to \(i\)
AmTran_Bus
  • AmTran_Bus
I thought the same thing until part B that wants Re(z^2). I know the right answer for it but cant get it.
ganeshie8
  • ganeshie8
z = x+2iy z^2 = (x+2iy)^2 = ?
AmTran_Bus
  • AmTran_Bus
But I thought it only wanted the real part. If you square all of it for this one, why did you only worry about the x in the other and not the imaginary 2iy in the other one?
ganeshie8
  • ganeshie8
because they are asking just the real part
AmTran_Bus
  • AmTran_Bus
Yea, so why is it not z^2, so z=x, so z^2 = x^2?
ganeshie8
  • ganeshie8
z is not x z is x+2iy
AmTran_Bus
  • AmTran_Bus
So what does the Re they specifify there mean?
ganeshie8
  • ganeshie8
to find the real part of z^2, first you need to find z^2
AmTran_Bus
  • AmTran_Bus
Ohh. Ohh. Ohh.
AmTran_Bus
  • AmTran_Bus
So can you let me work through it real quick?
ganeshie8
  • ganeshie8
sure
AmTran_Bus
  • AmTran_Bus
So I need (x+2iy)^2 first, which gives x^2+4iy-4y? is that right?
ganeshie8
  • ganeshie8
(x+2iy)^2 = x^2 + 4iy - 4y^2
AmTran_Bus
  • AmTran_Bus
How is the last part that?
AmTran_Bus
  • AmTran_Bus
I thought 2iy*2iy was 4i^2*y^2 Or -4y^3 Oh I understand.
ganeshie8
  • ganeshie8
(2iy)^2 = 4i^2y^2 = -4y^2
AmTran_Bus
  • AmTran_Bus
Oh so x^2-4y^2 is all the non imaginary stuff.
ganeshie8
  • ganeshie8
\(z = x+i2y\) \(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4y}\)
ganeshie8
  • ganeshie8
yes x^2-4y^2 is the real part
AmTran_Bus
  • AmTran_Bus
THANK YOU x1,000,000,000
ganeshie8
  • ganeshie8
np:)
AmTran_Bus
  • AmTran_Bus
Wait, why does the book sayx^2+4ixy-4y^2? Where did the x in the 4ixy come from? @ganeshie8
ganeshie8
  • ganeshie8
right, \(z = x+i2y\) \(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4xy}\)
ganeshie8
  • ganeshie8
we're just using the identity \[\large (\heartsuit + \spadesuit)^2~~ =~~ \heartsuit^2 + 2\heartsuit\spadesuit + \spadesuit^2 \]
AmTran_Bus
  • AmTran_Bus
|dw:1440213734204:dw| Sorry I'm so slow
ganeshie8
  • ganeshie8
|dw:1440213819660:dw|
AmTran_Bus
  • AmTran_Bus
yeppers. You came through again.
ganeshie8
  • ganeshie8
np

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