Complex Numbers Basics!

**Tutu** · November 26th 2012, 07:08 PM

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!

**Prove It** · November 26th 2012, 07:57 PM

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 $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ about the origin.

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

Then $\displaystyle \begin{align*} -\overline{z} = -r\,\textrm{cis}\,(-\theta) \end{align*}$ , 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 $\displaystyle \begin{align*} \pi \end{align*}$ .

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

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

$\displaystyle \begin{align*} z &= \sin{(\delta)} - i\cos{(\delta)} \\ &= \sin{(-\delta)} + i\cos{(-\delta)} \\ &= \cos{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} + i\sin{\left[ \frac{\pi}{2} - \left( -\delta \right) \right]} \\ &= \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \\ &= \textrm{cis}\left( \frac{\pi}{2} + \delta \right) \end{align*}$

**Tutu** · November 26th 2012, 08:24 PM

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!

**Prove It** · November 26th 2012, 08:37 PM

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.

$\displaystyle \begin{align*} \sin{(\delta)} - i\cos{(\delta)} &= -\left[ -\sin{(\delta)} + i\cos{(\delta)} \right] \\ &= - \left[ \sin{(-\delta)} + i\cos{(-\delta)} \right] \\ &= - \left\{ \cos{\left[ \frac{\pi}{2} - (-\delta) \right]} + i\sin{\left[ \frac{\pi}{2} - (-\delta) \right]} \right\} \\ &= - \left[ \cos{\left( \frac{\pi}{2} + \delta \right)} + i\sin{\left( \frac{\pi}{2} + \delta \right)} \right] \end{align*}$

**Tutu** · November 26th 2012, 08:51 PM

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) ) ?

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

Thank you so much!

**Prove It** · November 27th 2012, 07:46 PM

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) ) ?

to get the answer..
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 $\displaystyle \begin{align*} \cos{\left[ -\left( \frac{\pi}{2} + \delta \right) \right]} + i\sin{\left[ -\left( \frac{\pi}{2} + \delta \right) \right] } \end{align*}$ , not as you had it.

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

What we really have to do is remember that by negating an entire complex number, we are really rotating it by $\displaystyle \begin{align*} \pi \end{align*}$ , or if you like, reflecting it in both the real and imaginary axes. So if we subtract $\displaystyle \begin{align*} \pi \end{align*}$ from both angles, you should get your answer in the desired form

**Tutu** · November 28th 2012, 07:14 PM

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

**Prove It** · November 28th 2012, 11:49 PM

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 $\displaystyle \begin{align*} \textrm{cis}\left( \delta - \frac{\pi}{2} \right) \end{align*}$ . Like I said, to negate the complex number you need to rotate it by $\displaystyle \begin{align*} \pi \end{align*}$ units about the origin. Therefore, your angle should become $\displaystyle \begin{align*} \delta - \frac{\pi}{2} + \pi = \delta + \frac{\pi}{2} \end{align*}$ as required.

**Tutu** · November 29th 2012, 02:21 AM

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!

**Tutu** · December 5th 2012, 09:15 AM

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

**Plato** · December 5th 2012, 10:26 AM

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 $z=r\text{cis}(\theta)$ then $-z=r\text{cis}(\theta+\pi)$ . Notice that $|z|=|-z|=r$ and $-z$ is the reflection of $z$ through the origin, which is a half-turn about the origin.

**Deveno** · December 5th 2012, 11:11 AM

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).

**Tutu** · December 5th 2012, 11:38 AM

But the answer is cis( delta - pi/2 ), is it wrong?thank you'!!

**Plato** · December 5th 2012, 12:10 PM

Originally Posted by Tutu

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

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

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

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

The is the reason it works.

$\begin{align*}\overline{\text{cis}(\delta)}& =\overline{\cos(\delta)+i\sin(\delta)}\\ &=\cos(\delta)-i\sin(\delta) \\&=\cos(-\delta)+i\sin(-\delta) \\ &=\text{cis}(-\delta)\end{align*}$ .

**Deveno** · December 5th 2012, 12:14 PM

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.

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