Complex (1 Viewer)

[Lukybear]

Find three dif values of z for which.

z^3 = 1+i/sqrt 2

 

[gurmies]

 

[duckcowhybrid]

You might wanna add more steps mate. Eg General Soln to cis pi/4 yields pi/12 + 2kpi/3.

 

[The Nomad]

I think for the early questions, markers don't care much about working out as long as you give them the answer. Not for the simple questions, anyway.

 

[lychnobity]

You might wanna add more steps mate. Eg General Soln to cis pi/4 yields pi/12 + 2kpi/3.

It's in the 3rd line

 

[gurmies]

Pardon me if I was too vague. He could have posted with a follow-up question. As he didn't, I thought it sufficed.

 

[duckcowhybrid]

It's in the 3rd line

It's still a bit vague to me. I get it, but it seems a bit jumpy. But then again, I've just started complex so I really dunno how much working you need.

 

[gurmies]

Well, it all stems from the fact that roots are equally spaced out around a circle. If you take the first case, using De Moivre's Theorem, [cis(pi/4)]^(1/3) = cis(pi/12). This gives you the starting root. For the other two, just add and subtract 2kpi/3 [to maintain equal angular spacing]...

For the general case, if you want to find the nth root of a complex number, it's given by cis(@/n + 2kpi/n), k = 0, +-1, +-2, ...

 

[duckcowhybrid]

I knew that except the last line, which is a huge help to me. Thanks.

 

[gurmies]

I knew that except the last line, which is a huge help to me. Thanks.

Not a problem, sorry about the ambiguity.