If z = -3+4i....(related vectors)

[Raerin]

If z = -3+4i, determine the following related complex numbers

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."

 

[MarkFL]

Re: If z = -3+4i...

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?

 

[Raerin]

Re: If z = -3+4i...

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?

 

[MarkFL]

Re: If z = -3+4i...

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?

 

[Raerin]

Re: If z = -3+4i...

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?

Ahh, I see, so |z| would be 5.

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

 

[MarkFL]

Re: If z = -3+4i...

Yes, correct on both counts. (Sun)

So, what about parts b) and c)?

 

[Raerin]

Re: If z = -3+4i...

Yes, correct on both counts. (Sun)

So, what about parts b) and c)?

Okay, I understand everything now. Thanks for your help!

 

[I like Serena]

Re: If z = -3+4i...

Hi Raerin, welcome to MHB! :)

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

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}

 

[MarkFL]

Re: If z = -3+4i...

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}

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