Question

I don't think that I' doing this problem right so can someone please help/ check my work?

Answer 1

question: for any complex number z, show that:
\[a)Re(z)=\frac{(z+z^{*})}{2}\]
\[b)Im(z)=\frac{(z-z^{*})}{2i}\]

Answer 2

ok so I guess I make up two simple complex numbers. I'm confused why the function are labeled as Re(real) and Im(imaginary)

Answer 3

so I guess:
\[z=1+i\]
\[z^{*}=1-i\]

Answer 4

\[a)Re(z)=\frac{((1+i)+(1-i))}{2}\]
\[\frac{2}{2}=1\]

Answer 5

\[b)Im(z)=\frac{((1+i)-(1-i))}{2i}\]
\[\frac{1+i-1+i}{2i} \rightarrow \frac{2i}{2i}=1\]

Answer 6

You're correct, except you're proving it for the specific case where \(z = 1+i\). Do the exact same thing, except with \(z = a+bi\).

Answer 7

\(Re(z)\) is the real part of \(z\). For example, the real part of \(5+2i\) is \(5\). \(Im(z)\) is the imaginary part of \(z\). The imaginary part of \(5+2i\) is 2.

Answer 8

like this?
\[a)Re(z)=\frac{((a+bi)+(a-bi))}{2}=\frac{2a}{2}=a\]
\[b)Im(z)=\frac{((a+bi)-(a-bi))}{2i}=\frac{2bi}{2i}=b\]

Answer 9

Yep, exactly.

Answer 10

so when im asked, for example, give a geometric interpretation of why z x z* is a real number i can use the same concept?

Answer 11

or should i be plotting stuff

Answer 12

I'm not exactly sure how you would give a geometric interpretation of why it's true, but yes, you can use the same concept to show why it's real.

Answer 13

Yes, they probably want you plotting the numbers if they say geometric. The problem is \(z \ \cdot \ z*\) is multiplication, so you can't use vector addition.

Are you familiar with the polar form of complex numbers? \(z=Re^{i\theta}\)

Answer 14

euler's rule yes

Answer 15

so what have two different z and turn them into their polar forms then multiply them together?

Answer 16

Yes, so \(z=Re^{i\theta}\) and \(z^* = Re^{-i\theta}\). You can prove the latter by using Euler's formula.

Answer 17

This way, you can show that multiplying by a complex number with angle \(\theta\) is the same as rotating by \(\theta\) in the complex plane. That's your geometric interpretation.

Answer 18

so by using what you have:
\[Re^{i\theta}*Re^{-i\theta} \rightarrow Re^{0} \rightarrow R\]

Answer 19

Yes, except it's \(R^2\)

Answer 20

oh right and R^{2} is a real number

Answer 21

Yeah, and you can show that geometrically by plotting them with their angles labeled and then show how multiplying by \(z^*\) just "rotates \(z\) backwards" onto the real axis.

Answer 22

this is were i'm confused

Answer 23

Okay, so when you're plotting a complex number in polar form (\(z=Re^{i\theta}\)), you are basically plotting a vector with magnitude R at an angle \(\theta\) with the x-axis. Does this make sense?

Answer 24

i under stand how to plot them but what do you mean by it moving backwards

Answer 25

When you multiplied in polar form, you added \(-\theta\) to \(\theta\) in the exponent:
\[Re^{i\theta}∗Re^{−i\theta} = R^2e^{i(\theta - \theta)}\]

The new complex number is "rotated backwards" by theta. Do you see this in the exponent?

Answer 26

so the angle is just being subtracted by theta. i guess im getting confused because i can really see it on a graph. i just imagine that theta=> zero after subtraction so it should be on the x axis, but when i think of plotting re^(-itheta) i see it as the i value along negative y and theta being huge

Answer 27

would you mind helping my draw this so i understand

Answer 28

my thought