Finding polar and cartesian form for this power

[charmedbeauty]

Homework Statement

((-1+i)/(√2))^1002
find polar and cartesian form

Homework Equations

The Attempt at a Solution

So I started by finding |z|=1

and Arg(z)= arctan (-1) = 5pi/6

so 1^(1002)*e^(i*5pi/6)*1002 =1*e^(i*835pi)

but that's as far as I got because the answer says -i, but to get that term I need 1*e^-ipi/2

any thoughts??

 

[HallsofIvy]

Your main problem is that arctan(-1) is NOT 5\pi/6!

Calculate that again.

 

[charmedbeauty]

Your main problem is that arctan(-1) is NOT 5\pi/6!

Calculate that again.

I was assuming since the original eqn has a negative real part then I would add Pi/2 since the angle is measured from 0.
Although I am getting 3pi/4 now

 

[NewtonianAlch]

Well, yes (3pi)/4 is the angle you're looking for since this is in the second quadrant.

You have got |z|, and now you have the angle.

Put it into exponential form to the power of 1002, and simplify it to get the angle in terms of a principal argument.

 

[charmedbeauty]

Well, yes (3pi)/4 is the angle you're looking for since this is in the second quadrant.

You have got |z|, and now you have the angle.

Put it into exponential form to the power of 1002, and simplify it to get the angle in terms of a principal argument.

Ok so now I have 1^1002*(cos 3Pi/4+isin3Pi/4)

= 1*(0)=0 how do they get the minus i?

 

[NewtonianAlch]

Ok so now I have 1^1002*(cos 3Pi/4+isin3Pi/4)

= 1*(0)=0 how do they get the minus i?

How did you get 0 for (cos 3Pi/4 + isin 3Pi/4) ?

Also, you have forgotten to raise that to the power.

 

[charmedbeauty]

How did you get 0 for (cos 3Pi/4 + isin 3Pi/4) ?

Also, you have forgotten to raise that to the power.

isnt cos 3pi/4=-0.7...

and sin 3pi/4=0.7...

 

[NewtonianAlch]

isnt cos 3pi/4=-0.7...

and sin 3pi/4=0.7...

Yes, but you have a complex number there.

-0.707 +0.707i

You can't add them just like that.

What of the 1002? You need to account for that too.