# Complex Numbers Basics!

• Nov 26th 2012, 07:08 PM
Tutu
Complex Numbers Basics!
Hi I seem to not understand what the question is asking.

If z=2cis(delta),
write -z* in polar form.

I understand how to do this, my -z* was -2cos(delta) + i2sin(delta)
This is the second quadrant, and since
the coordinates of z are (2cos(delta), 2sin(delta))
the coordinates of -z* are (-2cos(delta), 2sin(delta)),
Can I can safely say that from z to-z*, it is a rotation of pi/2 about the origin?
Thus, my answer will be 2cis( delta + pi/2 ).

However, the answer is 2cis(pi - delta), which I understand because it still refers to the second quadrant, but I really wonder what is wrong with my answer. Looking at the coordinates, isn't it a pi/2 rotation? Unless my coordinates are wrong..

Sorry also, could you help me with this, Express sin(delta) - icos(delta) in polar form? The only thing I can gather is that the modulus equals to 1 and arctan(pi/2-delta).

Thank you so much!
• Nov 26th 2012, 07:57 PM
Prove It
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Hi I seem to not understand what the question is asking.

If z=2cis(delta),
write -z* in polar form.

I understand how to do this, my -z* was -2cos(delta) + i2sin(delta)
This is the second quadrant, and since
the coordinates of z are (2cos(delta), 2sin(delta))
the coordinates of -z* are (-2cos(delta), 2sin(delta)),
Can I can safely say that from z to-z*, it is a rotation of pi/2 about the origin?
Thus, my answer will be 2cis( delta + pi/2 ).

However, the answer is 2cis(pi - delta), which I understand because it still refers to the second quadrant, but I really wonder what is wrong with my answer. Looking at the coordinates, isn't it a pi/2 rotation? Unless my coordinates are wrong..

Sorry also, could you help me with this, Express sin(delta) - icos(delta) in polar form? The only thing I can gather is that the modulus equals to 1 and arctan(pi/2-delta).

Thank you so much!

No it's not a rotation of \displaystyle \displaystyle \begin{align*} \frac{\pi}{2} \end{align*} about the origin.

Notice that if \displaystyle \displaystyle \begin{align*} z = r\,\textrm{cis}\,(\theta) \end{align*}, then \displaystyle \displaystyle \begin{align*} \overline{z} = r\,\textrm{cis}\,(-\theta) \end{align*}. This represents a reflection about the Real axis.

Then \displaystyle \displaystyle \begin{align*} -\overline{z} = -r\,\textrm{cis}\,(-\theta) \end{align*}, which represents a complete reflection about both the Imaginary and Real axes. So by symmetry, in the second quadrant you end up with the same angle as in the first quadrant but swept out in a clockwise direction from the negative real axis. If we were to measure this angle from the positive real axis in the anticlockwise direction (as we normally do), then it's the same as subtracting that reference angle from \displaystyle \displaystyle \begin{align*} \pi \end{align*}.

Say our angle had been \displaystyle \displaystyle \begin{align*} \frac{\pi}{6} \end{align*}. When we perform the transformations to \displaystyle \displaystyle \begin{align*} z = r\,\textrm{cis}\,\frac{\pi}{6} \end{align*} to get \displaystyle \displaystyle \begin{align*} z = -r\,\textrm{cis}\,\frac{\pi}{6} \end{align*}, you should find you get \displaystyle \displaystyle \begin{align*} r\,\textrm{cis}\,\frac{5\pi}{6} \end{align*}, which is NOT a rotation of \displaystyle \displaystyle \begin{align*} \frac{\pi}{2} \end{align*} from \displaystyle \displaystyle \begin{align*} \frac{\pi}{6} \end{align*}.

To answer your second question, you should know that \displaystyle \displaystyle \begin{align*} \sin{(\theta)} = \cos{\left( \frac{\pi}{2} - \theta \right)} \end{align*} and \displaystyle \displaystyle \begin{align*} \cos{(\theta)} = \sin{\left( \frac{\pi}{2} -\theta \right)} \end{align*}. So

\displaystyle \displaystyle \begin{align*} z &= \sin{(\delta)} - i\cos{(\delta)} \\ &= \sin{(-\delta)} + i\cos{(-\delta)} \\ &= \cos{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} + i\sin{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} \\ &= \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \\ &= \textrm{cis}\left( \frac{\pi}{2} + \delta \right) \end{align*}
• Nov 26th 2012, 08:24 PM
Tutu
Re: Complex Numbers Basics!
Hi thank you so much! I understand the first part!

For the second question, does that mean that we can have sin(-delta) even though in the question sin(delta) - icos(delta), the sin is positive? I was stuck there since I thought we have to manipulate it in such a way that it fits the question.

Thank you so very much!
• Nov 26th 2012, 08:37 PM
Prove It
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Hi thank you so much! I understand the first part!

For the second question, does that mean that we can have sin(-delta) even though in the question sin(delta) - icos(delta), the sin is positive? I was stuck there since I thought we have to manipulate it in such a way that it fits the question.

Thank you so very much!

Actually I think I may have made a mistake in part 2.

\displaystyle \displaystyle \begin{align*} \sin{(\delta)} - i\cos{(\delta)} &= -\left[ -\sin{(\delta)} + i\cos{(\delta)} \right] \\ &= - \left[ \sin{(-\delta)} + i\cos{(-\delta)} \right] \\ &= - \left\{ \cos{\left[ \frac{\pi}{2} - (-\delta) \right]} + i\sin{\left[ \frac{\pi}{2} - (-\delta) \right]} \right\} \\ &= - \left[ \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \right] \end{align*}
• Nov 26th 2012, 08:51 PM
Tutu
Re: Complex Numbers Basics!
Thank you!

The answer is cis( delta - pi/2 ), so from your last working, all I have to do is to multiply the negative in
-( cos(pi/2 + delta) + isin(pi/2 + delta) )
= ( -cos(pi/2 + delta) - isin(pi/2 + delta) )
= ( cos(-pi/2 + delta) + isin(-pi/2 + delta) ) ?

then wouldn't it be ( cos(-pi/2 - delta) + isin(-pi/2 - delta) ) when I multiply the negative in?

Thank you so much!
• Nov 27th 2012, 07:46 PM
Prove It
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Thank you!

The answer is cis( delta - pi/2 ), so from your last working, all I have to do is to multiply the negative in
-( cos(pi/2 + delta) + isin(pi/2 + delta) )
= ( -cos(pi/2 + delta) - isin(pi/2 + delta) )
= ( cos(-pi/2 + delta) + isin(-pi/2 + delta) ) ?

then wouldn't it be ( cos(-pi/2 - delta) + isin(-pi/2 - delta) ) when I multiply the negative in?

Thank you so much!

First of all, be VERY careful with your brackets. You were trying to write \displaystyle \displaystyle \begin{align*} \cos{\left[ -\left( \frac{\pi}{2} + \delta \right) \right]} + i\sin{\left[ -\left( \frac{\pi}{2} + \delta \right) \right] } \end{align*}, not as you had it.

It's also incorrect. Remember that while \displaystyle \displaystyle \begin{align*} \sin{(-\theta)} = -\sin{(\theta)} \end{align*}, we have \displaystyle \displaystyle \begin{align*} \cos{(-\theta)} = \cos{(\theta)} \end{align*}, NOT \displaystyle \displaystyle \begin{align*} -\cos{(\theta)} \end{align*}.

What we really have to do is remember that by negating an entire complex number, we are really rotating it by \displaystyle \displaystyle \begin{align*} \pi \end{align*}, or if you like, reflecting it in both the real and imaginary axes. So if we subtract \displaystyle \displaystyle \begin{align*} \pi \end{align*} from both angles, you should get your answer in the desired form :)
• Nov 28th 2012, 07:14 PM
Tutu
Re: Complex Numbers Basics!
Thanks! That way, I'd get

-[cos(delta - (pi/2) + isin(delta - (pi/2)]

= - cis (delta - (pi/2)

But since modulus cannot be negative, it automatically becomes,

= cis (delta - (pi/2), is that right?

How do you know to negate the complex number? The answer negated the complex number, but the question did not indicate that such had to be done..is it intuitive?

Thank you so so m
• Nov 28th 2012, 11:49 PM
Prove It
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Thanks! That way, I'd get

-[cos(delta - (pi/2) + isin(delta - (pi/2)]

= - cis (delta - (pi/2)

But since modulus cannot be negative, it automatically becomes,

= cis (delta - (pi/2), is that right?

How do you know to negate the complex number? The answer negated the complex number, but the question did not indicate that such had to be done..is it intuitive?

Thank you so so m

No, it does not become \displaystyle \displaystyle \begin{align*} \textrm{cis}\left( \delta - \frac{\pi}{2} \right) \end{align*}. Like I said, to negate the complex number you need to rotate it by \displaystyle \displaystyle \begin{align*} \pi \end{align*} units about the origin. Therefore, your angle should become \displaystyle \displaystyle \begin{align*} \delta - \frac{\pi}{2} + \pi = \delta + \frac{\pi}{2} \end{align*} as required.
• Nov 29th 2012, 02:21 AM
Tutu
Re: Complex Numbers Basics!
But the answer is cis( delta - pi/2 ). To negate a complex number, you can choose to -pi or +pi right?

How do I know that I have to negate a complex number?

Thank you!
• Dec 5th 2012, 09:15 AM
Tutu
Re: Complex Numbers Basics!
Can I.ask, for the.negative in front of cis(pi/2+delta), can I just ignore it since cis.is never negative?
• Dec 5th 2012, 10:26 AM
Plato
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Can I.ask, for the.negative in front of cis(pi/2+delta), can I just ignore it since cis.is never negative?

Suppose that $\displaystyle z=r\text{cis}(\theta)$ then $\displaystyle -z=r\text{cis}(\theta+\pi)$. Notice that $\displaystyle |z|=|-z|=r$ and $\displaystyle -z$ is the reflection of $\displaystyle z$ through the origin, which is a half-turn about the origin.
• Dec 5th 2012, 11:11 AM
Deveno
Re: Complex Numbers Basics!
here is the way i look at it:

-(z*) = (-1)(z*) = (-1)*(z)* = (-z)*...it doesn't matter if we negate and then conjugate, or conjugate and then negate. suppose we conjugate first.

this turns cis(δ) into cis(-δ). now when we take the negative, we rotate by a half-turn, or π radians. this turns cis(-δ) into cis(π-δ).

or, we could negate first, turning cis(δ) into cis(π+δ). conjugating this gives us cis(-π-δ) = (1)(cis(-π-δ)) = cis(2π)cis(-π-δ) = cis(2π+(-π-δ)) = cis(π-δ), as before.

(using the fact that cis(x)cis(y) = cis(x+y), and that cis(2π) = cis(0) = 1).
• Dec 5th 2012, 11:38 AM
Tutu
Re: Complex Numbers Basics!
But the answer is cis( delta - pi/2 ), is it wrong?thank you'!!
• Dec 5th 2012, 12:10 PM
Plato
Re: Complex Numbers Basics!
Quote:

Originally Posted by Tutu
Hi I seem to not understand what the question is asking.
If $\displaystyle z=2\text{cis}(\delta)$, write $\displaystyle -\overline{z}$ in polar form.

That was in fact the original posting then the answer is:

$\displaystyle -\overline{z}=2\text{cis}(-\delta +\pi)$

Example:
$\displaystyle \begin{gathered} z = 2\text{cis}\left( {\frac{{2\pi }}{5}} \right) \hfill \\ \overline z = 2\text{cis}\left( { - \frac{{2\pi }}{5}} \right) \hfill \\ - \overline z = 2\text{cis}\left( {\frac{{3\pi }}{5}} \right) \hfill \\ \end{gathered}$

The is the reason it works.

\displaystyle \begin{align*}\overline{\text{cis}(\delta)}& =\overline{\cos(\delta)+i\sin(\delta)}\\ &=\cos(\delta)-i\sin(\delta) \\&=\cos(-\delta)+i\sin(-\delta) \\ &=\text{cis}(-\delta)\end{align*}.
• Dec 5th 2012, 12:14 PM
Deveno
Re: Complex Numbers Basics!
well, i *have* been know to make mistakes. let's pick a value for delta, and see what happens: we'll use δ = π/3.

so z = cis(π/3) = cos(π/3) + i sin(π/3) = 1/2 + i(√3/2).

then z* = 1/2 - i(√3/2), and -z* = -1/2 + i(√3/2). by inspection, we see this is cis(2π/3) (a primitive cube root of unity) (note it is in the second quadrant)

so which formula is right: π-δ, or δ-π/2?

π-δ = π - π/3 = 2π/3. this agrees with what we found.

δ-π/2 = π/3 - π/2 = -π/6, which is in the fourth quadrant. this is not right. re-read your ORIGINAL post. i stand by what i posted.