Complex numbers help

[pinklemonade]

The question is:
Let w=5cis and z=-i
a.) Find Arg(z)
b.) Use the result of part a. to find Arg(w^2z)

Thank you!

[keltingmeith]

The question is:
Let w=5cis and z=-i
a.) Find Arg(z)
b.) Use the result of part a. to find Arg(w^2z)

Thank you!

Protip: maybe put all the parts into TeX instead of just select parts - the size differences make it harder to read...

a) The argument is just the angle it makes with the positive x-axis, so we create a triangle and solve for the angle - or, using a very common formula,

b) We know that if u and v are complex numbers, then . So, this means that

Of course, this isn't in our preferred domain - so, we take off 2pi, getting

[IndefatigableLover]

Quick question EulerFan101 but is there something wrong in your working out since I got as my answer for part b (using a different method)..

EDIT:

Using De Moivre's Theorem, we can find w^2 and then make it into polar form:


^Changing this into polar form will yield us:

(which doesn't look nice)

However, when we find the angle between the two of these using the same method as EulerFan101 did in Part A, these two numbers cancel quite nicely where we will get:



(due to the quadrant we are in when plotting this out on an argand diagram).

From there we add both argument values together to get our answer (that is which will yield us

[keltingmeith]

Quick question EulerFan101 but is there something wrong in your working out since I got as my answer for part b (using a different method)..

EDIT:

Using De Moivre's Theorem, we can find w^2 and then make it into polar form:


^Changing this into polar form will yield us:

(which doesn't look nice)

However, when we find the angle between the two of these using the same method as EulerFan101 did in Part A, these two numbers cancel quite nicely where we will get:



(due to the quadrant we are in when plotting this out on an argand diagram).

From there we add both argument values together to get our answer (that is which will yield us

Yeah, I think you've made a silly mistake. Your angle for w^2 is wrong - , not . Otherwise, your application of De Moivre's Theorem is perfect - and something cool/important to note is that while our methods seem different, we're actually doing the same thing, whereas you're just looking at a different side of it.

Also, your finding of the angle after De Moivre's isn't wrong, but definitely unnecessary - you've converted from polar form to cartesian form, so that you can find the angle - which is defined in polar form... It's like taking a cooked sausage roll, then letting it cool down so that you can re-heat it in the microwave.

[pinklemonade]

Hey I also have got another question that I don't understand...

Show that 2-i is a solution of the equation

When I factorise that equation I get (z+1)(z-1)(z-2+i) so I don't know what I'm doing wrong!

[lzxnl]

Protip: maybe put all the parts into TeX instead of just select parts - the size differences make it harder to read...

a) The argument is just the angle it makes with the positive x-axis, so we create a triangle and solve for the angle - or, using a very common formula,

Objection: that formula doesn't work if the real part of your complex number isn't positive. It works here, but be wary if you blindly use that formula.

Also you had a typo in your working there's a 21 where there should be a 12

Hey I also have got another question that I don't understand...

Show that 2-i is a solution of the equation

When I factorise that equation I get (z+1)(z-1)(z-2+i) so I don't know what I'm doing wrong!

You've actually done nothing wrong, except you've forgotten to sub in z = 2-i to show that the polynomial is zero there.

[IndefatigableLover]

Yeah, I think you've made a silly mistake. Your angle for w^2 is wrong - , not . Otherwise, your application of De Moivre's Theorem is perfect - and something cool/important to note is that while our methods seem different, we're actually doing the same thing, whereas you're just looking at a different side of it.

Also, your finding of the angle after De Moivre's isn't wrong, but definitely unnecessary - you've converted from polar form to cartesian form, so that you can find the angle - which is defined in polar form... It's like taking a cooked sausage roll, then letting it cool down so that you can re-heat it in the microwave.

Ah damn I redid my angle thing and I accidentally subtracted by rather than which led me to get that answer (if I did it by then I would have gotten the same thing)! Thanks for the clear up

[pinklemonade]

For any complex number z, the complex number is found by
A. reflecting z in the Im(z) axis
B. reflecting z in the Re(z) axis
C. reflecting z in the line Im(z)=Re(z)
D. Rotating z through in a clockwise direction about the origin
E. Rotating z through in an anti-clockwise direction about the origin

I don't know how to do this one at all

[nhmn0301]

For any complex number z, the complex number is found by
A. reflecting z in the Im(z) axis
B. reflecting z in the Re(z) axis
C. reflecting z in the line Im(z)=Re(z)
D. Rotating z through in a clockwise direction about the origin
E. Rotating z through in an anti-clockwise direction about the origin

I don't know how to do this one at all

so i^5 = i
u= iz, which is similar to rotate 90 degree anticlockwise. I would choose E.

[pinklemonade]

If , then z is equal to
 

[IndefatigableLover]

If , then z is equal to
 

I got A. as my answer! My method is a bit unconventional (I'm sure one of the Math's gurus can show you a better approach since mine is more of an intuitive approach).

So you know from De Moivre's Theorem that is the same as

(since the 2 at the front will become positive when squared). Essentially z=.... is the same as z^1 so if you replace all the '2' for '1', you will get:


From there you just do both cases and you should end up with A. as your answer

[lzxnl]

I use a method that is intuitively simple but expressing everything makes it messy.

As I'm finding the square roots, I know that all I have to do is square root the modulus, halve the argument and perhaps add /subtract pi from the argument. Here, because our required answer is in Cartesian form, I just go 'find one square root and then put a minus sign in front'.
So, one square root is obviously 2 cis 2pi/3  = -1 + sqrt3 i. The other one has to then be 1 - sqrt3 i, so A.

To adapt this for other powers, if I wanted fifth roots, I would just fifth root the argument, divide the argument by five and add multiples of 2pi/5 to the argument, using the principle that the nth roots of any complex number are evenly spaced out around a circle.

I got A. as my answer! My method is a bit unconventional (I'm sure one of the Math's gurus can show you a better approach since mine is more of an intuitive approach).

So you know from De Moivre's Theorem that is the same as

(since the 2 at the front will become positive when squared). Essentially z=.... is the same as z^1 so if you replace all the '2' for '1', you will get:


From there you just do both cases and you should end up with A. as your answer

Only problem with your answer is that z^2 is not +- something. You've jumped the gun a bit.

[keltingmeith]

Only problem with your answer is that z^2 is not +- something. You've jumped the gun a bit.

It does work if you change the angle to remove the negative (i.e subtract pi from the angle and make the negative a positive).

[lzxnl]

It does work if you change the angle to remove the negative (i.e subtract pi from the angle and make the negative a positive).

You misunderstood my point. It's like saying when square rooting z^2 = 1, you solve z^2 = +-1
Which, of course, gives you the solutions of z^4 = 1 and is different to solving z^2 = 1

In your question, z^2 is one fixed complex number. A +- sign does not mean one fixed complex number.