Complex number into a trigonometric form

L

Loki123

Full Member
Joined
Sep 22, 2021
Messages
790
Turn z=1-cosa+isina into a trigonometric form.
I attached my work but I think it's pretty useless. I have no idea how they got the correct solution which is on the third picture. IMG_20220305_124208.jpgIMG_20220305_124219.jpgIMG_20220305_124243.jpg
 
BigBeachBanana said:
I checked with Wolfram Alpha, it doesn't agree that the answer is correct.
[math] z=1-\cos(\alpha)+i\sin(\alpha)\\ =2\left(\frac{1-\cos(\alpha)}{2}\right)+i\sin(\alpha)\\ =2\sin^2\left(\frac{\alpha}{2}\right)+ i\cdot 2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\\ =2\sin\left(\frac{\alpha}{2}\right)\left[\sin\left(\frac{\alpha}{2}\right) +i\cos\left(\frac{\alpha}{2}\right)\right]\\ =2\sin\left(\frac{\alpha}{2}\right)\left[\cos\left(\frac{\pi}{2}-\frac{\alpha}{2}\right) +i\sin\left(\frac{\pi}{2}-\frac{\alpha}{2}\right) \right] [/math]
Click to expand...
Isn't that "almost" same as the third image (presumably the answer) in the OP:

1646529176193.png

I am saying that with my tongue firmly implanted in my cheek.
 
Subhotosh Khan said:
Why don't you post a photograph of the assignment ?
Click to expand...
Perhaps we should rather be asking: "Why won't you post a photograph of the assignment ?"!
Loki has repeatedly been asked to provide 'original material' but has steadfastly refused to do so!
And every thread (that I have seen) where Loki is the OP seems to end with no "resolution" and a fresh one is started that is replete with further ambiguity!
Maybe it is time to resist providing (unavoidably) speculative 'help' on Loki's posts until s/he is prepared to give us what we ask for?
 
The Highlander said:
Perhaps we should rather be asking: "Why won't you post a photograph of the assignment ?"!
Loki has repeatedly been asked to provide 'original material' but has steadfastly refused to do so!
And every thread (that I have seen) where Loki is the OP seems to end with no "resolution" and a fresh one is started that is replete with further ambiguity!
Maybe it is time to resist providing (unavoidably) speculative 'help' on Loki's posts until s/he is prepared to give us what we ask for?
Click to expand...
I am not avoiding, I either don't have it. I get these written. Or the problem is solved by the time you ask.
 
Loki123 said:
I am not avoiding, I either don't have it. I get these written. Or the problem is solved by the time you ask.
Click to expand...
Then you should state in your OP:

I do NOT have the original problem statement and this statement was given to me by my instructor or another student.​
​
And​
​
if you found the solution prior to our "complain", you should quickly post the "corrected" solution.​
 
Subhotosh Khan said:
Then you should state in your OP:

I do NOT have the original problem statement and this statement was given to me by my instructor or another student.​
​
And​
​
if you found the solution prior to our "complain", you should quickly post the "corrected" solution.​
Click to expand...
I will from now on. Thank you for understanding.
 
Suppose that [imath]z=x+{\bf \mathcal{i}}y[/imath] such that [imath]x\cdot y\ne 0[/imath] then
[imath]\arg(z)=\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y>0\\[/imath]
[imath]\arg(z)=-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y<0\\[/imath]
[imath]\arg(z)=\pi-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y>0\\[/imath]
[imath]\arg(z)=-\pi +\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y<0\\[/imath]
[imath][/imath][imath][/imath][imath][/imath]
 
pka said:
Suppose that [imath]z=x+{\bf \mathcal{i}}y[/imath] such that [imath]x\cdot y\ne 0[/imath] then
[imath]\arg(z)=\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y>0\\[/imath]
[imath]\arg(z)=-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y<0\\[/imath]
[imath]\arg(z)=\pi-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y>0\\[/imath]
[imath]\arg(z)=-\pi +\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y<0\\[/imath]
[imath][/imath][imath][/imath][imath][/imath]
Click to expand...
I am not familiar with that.
 
pka said:
Suppose that z=x+iyz=x+{\bf \mathcal{i}}yz=x+iy such that x⋅y≠0x\cdot y\ne 0x⋅y=0 then
arg⁡(z)=arctan⁡(∣yx∣) if x>0 & y>0\arg(z)=\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y>0\\arg(z)=arctan(∣∣∣∣xy∣∣∣∣) if x>0 & y>0
arg⁡(z)=−arctan⁡(∣yx∣) if x>0 & y<0\arg(z)=-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x>0~\&~y<0\\arg(z)=−arctan(∣∣∣∣xy∣∣∣∣) if x>0 & y<0
arg⁡(z)=π−arctan⁡(∣yx∣) if x<0 & y>0\arg(z)=\pi-\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y>0\\arg(z)=π−arctan(∣∣∣∣xy∣∣∣∣) if x<0 & y>0
arg⁡(z)=−π+arctan⁡(∣yx∣) if x<0 & y<0\arg(z)=-\pi +\arctan\left(\left|\dfrac{y}{x}\right|\right)\text{ if }x<0~\&~y<0\\arg(z)=−π+arctan(∣∣∣∣xy∣∣∣∣) if x<0 & y<0
Click to expand...
Loki123 said:
I am not familiar with that.
Click to expand...
Make yourself very familiar with what pka posted. It will be extremely helpful in engineering.
 
Loki123 said:
i believe it is a little early for that
Click to expand...
Not at all! :mad:
It's something you should already know! :unsure:
It's just an expression of the fact that, depending on where the number lies in the complex plane, the Argument must obey the All, Sin, Tan, Cos "rule". 8-)
 
Last edited:
Loki123 said:
i believe it is a little early for that
Click to expand...
Why in the world would say that? You were told to change a complex number form rectangular to polar form.
To do that one must know the argument of the complex number.
 
You must log in or register to reply here.
Top