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#### Raerin

##### Member

- Oct 7, 2013

- 46

a) vector z

b) 3(vector z)

c) 1/z

d) 1/(vector z)

e) |z|

f) |vector z|

g) (vector z)/(|z|^2)

I'm not sure if it's a vector, but the z has a short line above it when I say "vector z."

- Thread starter Raerin
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- Thread starter
- #1

- Oct 7, 2013

- 46

a) vector z

b) 3(vector z)

c) 1/z

d) 1/(vector z)

e) |z|

f) |vector z|

g) (vector z)/(|z|^2)

I'm not sure if it's a vector, but the z has a short line above it when I say "vector z."

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- #3

- Oct 7, 2013

- 46

The line over a complex number refers to its conjugate. If $z=a+bi$, then $\overline{z}=a-bi$. Are you familiar with the other notations?

Nope but |z| refers to the length of it? So it'll be the radius of a circle?

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- Oct 7, 2013

- 46

Ahh, I see, so |z| would be 5.Yes, $|z|$ refers to the magnitude, which is given by:

\(\displaystyle |z|=\sqrt{a^2+b^2}\)

So, what do you find for the magnitude of the given complex number?

Also, when 1 is divided by z, is it the same as dividing 1 with...5, let's say?

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- Oct 7, 2013

- 46

Okay, I understand everything now. Thanks for your help!Yes, correct on both counts.

So, what about parts b) and c)?

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- #8

- Mar 5, 2012

- 9,417

Hi Raerin, welcome to MHB!

I'd like to add a bit of nuance here.Also, when 1 is divided by z, is it the same as dividing 1 with...5, let's say?

\begin{aligned}

\frac 1 z &= \frac 1 {-3+4i} \\

&= \frac 1 {-3+4i} \cdot \frac {-3-4i} {-3-4i} \\

&= \frac {-3-4i} {(-3+4i)(-3-4i)} \\

&= \frac {-3-4i}{(-3)^2-(4i)^2} \\

&= \frac {-3-4i}{9+16} \\

&= \frac 1 {25} (-3-4i)

\end{aligned}

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Yes, I was mistakenly referring to \(\displaystyle \frac{1}{|z|}\). Good catch!Hi Raerin, welcome to MHB!

I'd like to add a bit of nuance here.

\begin{aligned}

\frac 1 z &= \frac 1 {-3+4i} \\

&= \frac 1 {-3+4i} \cdot \frac {-3-4i} {-3-4i} \\

&= \frac {-3+4i} {(-3+4i)(-3-4i)} \\

&= \frac {-3-4i}{(-3)^2-(4i)^2} \\

&= \frac {-3-4i}{9+16} \\

&= \frac 1 {25} (-3-4i)

\end{aligned}