Question

How do you represent this equation with de moivre's EXPONENTIAL formula??

Answer 1

\[u=i \sqrt{z}\]

Answer 2

So assuming that z is some complex function that we can write in the form:
\[\Large\rm z=r(\cos \theta+\mathcal i \sin \theta)\]
Err I guess you specifically wrote EXPONENTIAL so we'll write it in exponential form:\[\Large\rm z=r e^{\mathcal i \theta}\]

Plugging this in gives us:\[\Large\rm u=\mathcal i \left(r e^{\mathcal i \theta}\right)^{1/2}\]Which simplifies to, \[\Large\rm u=\mathcal i r^{1/2} e^{\mathcal i \theta/2}\]Mmmm I hope I did that right....
Are you told anything specific about z?

Answer 3

Ok

Answer 4

But there is something that I don't understand

Answer 5

On my book it says that the angles that i find are four. How do I find them.
It says that they are\[\frac{ \pi }{ 2 }<\theta<\frac{ 3\pi }{ 2 }\] and \[\frac{ \pi }{ 2 }+\pi<\theta<\frac{ 3\pi }{ 2 }+\pi\]

Answer 6

I really don't get it. But the process is correct!

Answer 7

Thanks

Answer 8

I'm confused.. these
\[\frac{ \pi }{ 2 }<\theta<\frac{ 3\pi }{ 2 }\]and\[\frac{ \pi }{ 2 }+\pi<\theta<\frac{ 3\pi }{ 2 }+\pi\]Are not angles, they're intervals.
So I'm not really sure what they're looking for.. hmmm

Answer 9

Um the problem says to find the solutions and represent them on the graph

Answer 10

And these should be the intervals just like you said. I am not english so I might use the wrong terms...

Answer 11

Oh oh oh, ok let's try this...

Answer 12

Does this make sense?\[\Large\rm \mathcal i=e^{\mathcal i \pi/2}\]

Answer 13

\[\Large\rm u=\color{green}{\mathcal i} \left(r e^{\mathcal i \theta}\right)^{1/2}=\color{green}{e^{\mathcal i \pi/2}} \left(r e^{\mathcal i \theta}\right)^{1/2}\]We should probably write our angle like this:\[\Large\rm u=e^{\mathcal i \pi/2} \left(r e^{\mathcal i( \theta+2k \pi)}\right)^{1/2},\qquad k=0,~1,~2,...\]
The idea is, we have some angle, but if we add or subtract 2pi from that angle, we land at the same location, yes?

Since we're taking the `2nd root`,
There will be only 2 unique roots, and the `first 2 values of k` give them to us.\[\Large\rm u=e^{\mathcal i \pi/2} \sqrt{r} ~e^{\mathcal i( \theta/2+k\pi)},\qquad k=0,~1\]So let's bring in the e^{i pi/2},\[\Large\rm u=\sqrt{r} ~e^{\mathcal i( \theta/2+k\pi+\pi/2)},\qquad k=0,~1\]So the 2 k values are giving us:\[\Large\rm u_0=\sqrt{r} ~e^{\mathcal i( \theta/2+\pi/2)}\]\[\Large\rm u_1=\sqrt{r} ~e^{\mathcal i( \theta/2+3\pi/2)}\]

Answer 14

And then ummmm >.< hmm

Answer 15

Maybe I'm over complicating this... ugh..

Answer 16

\(\large u=i (z^{\frac{1}{2}})\)
\(\large z=r e^{ i\theta}\)
\(\frac{ \pi }{ 2 }<\theta<\frac{ 3\pi }{ 2 }\)
this is only what ou got ?
does it said that z lie on unit circle ?

Answer 17

would u plz type the whole Qn ?

Answer 18

I thought I had something, but it still didn't make sense with the specific intervals your book gives.

Answer 19

It's not my fault xD

Answer 20

only if u type the whole Qn , it might make some sense

Answer 21

The text says to write the exponential formula for that equation and to represent it on the graph.

Answer 22

Perhaps if we tried using De Moivre's formula we might get somewhere. Let \(z=r(\cos(\theta)+i\sin(\theta))\). Then\[z^{1/2}=r^{1/2}(\cos(\theta/2)+i\sin(\theta/2))\]Multiplying by \(i\), we get\[iz^{1/2}=r^{1/2}(i\cos(\theta/2)-\sin(\theta/2)).\]This doesn't seem to be helping...

Answer 23

mmm well assume its on unit circle , then it would be helpfull

Answer 24

or |z| <=2 mm anything might work

Answer 25

or
u^2=-z

Answer 26

u=iz=(x+iy)
=-y+xi
>.>

Answer 27

I think @zepdrix was on the right track though. In general square roots have two possible answers. One root should have that \(\theta\) falls in the first interval you gave, and the second falls in the other interval.