Question

Complex numbers

Answer 1

If Z=x+2iy, then find Re(z*)

Where z=x+iy is the formula and x=Re(z) and y=IM(z)

Answer 2

Re= real , IM=imag.

Answer 3

* is the conjugate

Answer 4

\(z^*\) is obtained by reflecting the complex number over real axis
it is called the conjugate of \(z\)

Answer 5

when you reflect something over real axis,
notice that the real component is not changed

Answer 6

Here is how it says to do it: Re(z*) =Re(x-2iy) =x. Can you help me understand it that way?

Answer 7

just enter the part that is not attached to \(i\)

Answer 8

I thought the same thing until part B that wants Re(z^2). I know the right answer for it but cant get it.

Answer 9

z = x+2iy

z^2 = (x+2iy)^2 = ?

Answer 10

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?

Answer 11

because they are asking just the real part

Answer 12

Yea, so why is it not z^2, so z=x, so z^2 = x^2?

Answer 13

z is not x

z is x+2iy

Answer 14

So what does the Re they specifify there mean?

Answer 15

to find the real part of z^2,
first you need to find z^2

Answer 16

Ohh. Ohh. Ohh.

Answer 17

So can you let me work through it real quick?

Answer 18

sure

Answer 19

So I need (x+2iy)^2 first, which gives

x^2+4iy-4y? is that right?

Answer 20

(x+2iy)^2 = x^2 + 4iy - 4y^2

Answer 21

How is the last part that?

Answer 22

I thought 2iy*2iy was 4i^2*y^2
Or -4y^3
Oh I understand.

Answer 23

(2iy)^2 = 4i^2y^2 = -4y^2

Answer 24

Oh so x^2-4y^2 is all the non imaginary stuff.

Answer 25

\(z = x+i2y\)

\(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4y}\)

Answer 26

yes x^2-4y^2 is the real part

Answer 27

THANK YOU x1,000,000,000

Answer 28

np:)

Answer 29

Wait, why does the book sayx^2+4ixy-4y^2? Where did the x in the 4ixy come from? @ganeshie8

Answer 30

right,

\(z = x+i2y\)

\(z^2 = (x+2iy)^2 = \color{red}{x^2-4y^2} + i\color{purple}{4xy}\)

Answer 31

we're just using the identity
\[\large (\heartsuit + \spadesuit)^2~~ =~~ \heartsuit^2 + 2\heartsuit\spadesuit + \spadesuit^2 \]

Answer 32

Sorry I'm so slow