What Is the Argument of j to the Fourth Power?

[aliz_khanz]

okie ... one attempt as i see it ...

argument of j is pie/2 ... so argument of j^2 will be pie square/ 4 and so on ...

is it right?

 

[jhae2.718]

No. Here's a hint: what is j⁴ equivalent to?

For some number $z=x+y\mathrm{j} \in \mathbb{C}$, $\mathrm{arg}(z) \equiv \arctan(\frac{y}{x})$.

 

[aliz_khanz]

Dear Jhae2.718 ... can you please explain it in polar form?

for example...

z= 1(cos pie/2 +jsinpie/2) is the argument of j ...i don't know what to do for j^4

 

[jhae2.718]

For polar form, the argument is just $\theta$.

Keep in mind that we're using $z = r\cos(\theta)+\mathrm{j}r\sin(\theta)$.

For jⁿ, you'll want to simplify a bit. Recall:

n      j[sup]n[/sup] 
-----------------------
1      j
2     j*j = - 1
3     j*j[sup]2[/sup] = -j
4     j*j[sup]3[/sup] = 1
.
.
.

 

[aliz_khanz]

so i take from your explanation that ...

z= rcos (theta) + rsin (theta) *since j^4=1

 

[jhae2.718]

j⁴ =1, which means that the imaginary part of z is zero. What value of $\theta$ makes $\sin(\theta) \to 0$?

 

[aliz_khanz]

sin 0 = 0

can you please provide me the final answer of this bro? :)

 

[jhae2.718]

arg(j^4) = 0, since j^4 is on the positive real axis and the argument is the angle between the complex no. and the positive real axis.

 

[aliz_khanz]

shouldnt it be 1? because sin theta goes to zero , but cos theta goes to 1 !

 

[jhae2.718]

j^4 is 1, but the argument of j^4 is zero. As a general rule, positive values on the real axis have an arg of 0 and positive imaginary values have an arg of pi/2.

Remember, in polar coordinates the argument is just the angle theta!

 

[aliz_khanz]

Thankyou so so so so so so times infinity much ! Really ... ! May God Bless You with A billion dollar or equivalent ! :)

 

[jhae2.718]

No problem

If you want to read more about complex arguments, you can look at the http://en.wikipedia.org/wiki/Argument_(complex_analysis)" . It gets a bit technical, but the graphs should help.

 

[HallsofIvy]

okie ... one attempt as i see it ...

argument of j is pie/2 ... so argument of j^2 will be pie square/ 4 and so on ...

is it right?

No, squaring a number multiplies the argument by 2- it does not square it! The argument of j^4 is 4(\pi/2)= 2\pi (or 0 since adding or subtracting any multiple of 2pi won't change the answer.)

Of, same same thing, j^2= -1 so j^4= (j^2)(j^2)= (-1)(-1)= 1 which has argument 0.