2

$z=4\mathrm{e}^{i1.7}$

Write the following numbers in the polar form $r\mathrm{e}^{i\phi}$:

a) $\bar{z}$

b) $\displaystyle \frac{1}{z}$

I need $r$ and $\phi$. I already have $r$ for the both of these (a is $4$, b is $.25$) but I'm not sure what $\phi$ is for either of them. I thought it would be $1.7$ for a and $1/1.7$ for b, but those are wrong. I'm also not sure what the bar over the $z$ in part (a) is.

Lutz Lehmann
  • 126,666
calicojack
  • 41
  • 1
    Use that $z \times \overline{z} = |z|^2,~$ which is a real number. Further, use (as an immediate consequence of the definition of a conjugate) that $\overline{re^{i\theta}} = re^{-i\theta}.$ – user2661923 Aug 26 '21 at 02:45

1 Answers1

1

The bar signifies that $\bar{z}$ is the conjugate of the complex number $z$. Conjugates of complex numbers are complex numbers with the same real part, but with the negative of their imaginary parts.

For example, if $z=3 + 4i$, $\bar{z}$ is $3-4i$.

That said, if you also know how complex numbers are represented geometrically, you should be able to use Euler's formula to figure out what the angle is for a.

For b, you take take the reciprocal of z;

$\frac{1}{z}=\frac{1}{4e^{i1.7}}$

You're right about the value of r here. What do you think the value of $\phi$ would be if you simplify it into the standard polar form you've written of in your post?

Besides the links above, reading up on the basics of complex numbers should help.

Cheers!

harry
  • 1,076
  • I thought the phi answer in (b) would then be -1.7, but that wasn't correct either. I thought -1.7 would be the answer for (a) as well, but they were both wrong. I'm honestly lost here, my instructor hasn't gone over complex numbers yet. – calicojack Aug 27 '21 at 04:48
  • @calicojack: I checked it with someone and I'm still pretty sure that's how it works out. Your book seems incorrect. – harry Aug 28 '21 at 13:14