z=e^(2+(pi/4)i): Find |z|, Re z, Im z, z-bar in Cart. form

[cheffy]

Say that

\(\displaystyle \L z = e^{2 + i\frac{\pi }{4}}\)

Find |z|, Re z, Im z, and z bar in Cartesian form.

I don't even know what Re z or Im z mean and I'm not quite sure how to start the rest. Help!

Thanks

 

[daon]

If \(\displaystyle z = a+bi\) then \(\displaystyle Re\(z\) = a\) and \(\displaystyle Im \(z\)\) = b.

You are given \(\displaystyle z=e^{2+i\frac{\pi}{4}}\).

Using Euler's famous formula: \(\displaystyle e^{2+i\frac{\pi}{4}} \,\, = \,\, e^2\( cos(\frac{\pi}{4}) + i sin(\frac{\pi}{4})\) \,\, = \,\,e^2 cos(\frac{\pi}{4}) + i e^2sin(\frac{\pi}{4}) \,\,\)

Then \(\displaystyle |z| = \sqrt{ \(Re(z)\)^2 + \(Im(z)\)^2}\).

-daon

 

[stapel]

I don't even know what Re z or Im z mean and I'm not quite sure how to start the rest.

This should have been covered extensively in class...? Since it wasn't (?) and since we cannot reasonably provide the necessary instruction, you might want to try some online lessons, such as:

. . . . .Harvey Mudd College: Complex Numbers

The above lesson explains, in part, that a complex number z = a + bi is the sum of a real part, Re(z) = a, and an imaginary part, Im(z) = b. For instance, z = 2 + 3i is the sum of 2 and 3i, and Re(z) = 2 and Im(z) = 3. The modulus of z, |z|, is given by a² + b². The other part, z-bar, is just the complex conjugate: for z = a + bi, z-bar = a - bi.

Hope that helps a bit.

Eliz.

 

[cheffy]

oh oops. I just didn't know what the Re and the Im stood for since we called them differently in class and in the book.

Thanks!

 

[pka]

This makes an interesting proof.
\(\displaystyle \L\left| {e^z } \right| = e^{{{\rm Re}\nolimits} (z)}\)

 

[cheffy]

i'm getting that |z| = 0. Does that mean that z bar would be 0 too? Or do I have |z| wrong?

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
e^z = e^{x + yi} & = & e^x \left[ {\cos \left( y \right) + i\sin \left( y \right)} \right] \\
& = & \left[ {e^x \cos \left( y \right) + e^x i\sin \left( y \right)} \right] \\
\end{array}.\)

\(\displaystyle \L\begin{array}{rcl}
\left| {e^z } \right| & = & \left| {e^x \cos \left( y \right) + e^x i\sin \left( y \right)} \right| \\
& = & \sqrt {\left( {e^x \cos \left( y \right)} \right)^2 + \left( {e^x \sin \left( y \right)} \right)^2 } \\
& = & e^x \sqrt {\left( {\cos \left( y \right)} \right)^2 + \left( {\sin \left( y \right)} \right)^2 } \\
& = & e^x \\
& = & e^{{{\rm Re}\nolimits} (z)} \\
\end{array}\)

 

[daon]

i'm getting that |z| = 0. Does that mean that z bar would be 0 too? Or do I have |z| wrong?

As pka demonstrated, you answer for the norm should not be 0. Additionally, there should be a theorem you learned/will learn that says that |z|=0 if and only if z=0. This is called the "positive definite" property.