Complex number help! (2 Viewers)

Glyde · Oct 16, 2015

Express z1+z2 in mod-art form if z1=1+(Squareroot3)i and z2=(Squareroot3)+i This is probably really basic but I just started complex numbers this week

leehuan · Oct 16, 2015

z1 + z2 = 1+sqrt3*i + sqrt3+i

= (1+sqrt3) + (1+sqrt3)i

You then find the modulus to be sqrt2 * (1+sqrt3) and argument to be pi/4

So it's (1+sqrt3)sqrt2cis(pi/4)

InteGrand · Oct 16, 2015

Glyde said:
Express z1+z2 in mod-art form if z1=1+(Squareroot3)i and z2=(Squareroot3)+i This is probably really basic but I just started complex numbers this week

$Given: $z_1 = 1+\sqrt{3}i$ and $z_2 = \sqrt{3}+i$.$

$$\therefore z_1 + z_2 = 1+\sqrt{3}i + \sqrt{3}+i = \left(1+\sqrt{3} \right)+i\left( 1 + \sqrt{3}\right)i$ (add the real and imaginary parts respectively to get the sum of two complex numbers)$

$So $z_1 + z_2 = \left(1+\sqrt{3} \right) (1+i)$. We have$

$$=\mathrm{arg} (z_1 + z_2 ) = \mathrm{arg}\left[ \left(1+\sqrt{3} \right) (1+i)\right]$$

$$=\mathrm{arg}(1+i)$ (the positive constant in front of $(1+i)$ does not affect the arg, since the angle is the same)$

$= \tan^{-1} \frac{1}{1} =\frac{\pi}{4}$

$, and $|z_1 + z_2|= \left | \left(1+\sqrt{3} \right) (1+i) \right |=\left |1+\sqrt{3} \right | \times |1+i| = \left(1+\sqrt{3} \right) \times \sqrt{1^2 + 1^2}= \left(1+\sqrt{3} \right)\sqrt{2}$.$

$Now that we now the modulus and argument of $z_1 + z_2$, we have $z_1 + z_2 = \left(1+\sqrt{3} \right)\sqrt{2} \left[\cos \frac{\pi}{4} + i\sin \frac{\pi}{4} \right] $.$

Ambility · Oct 16, 2015

Hey! I just started complex numbers this week as well! I might be wrong, but I'll give it a try.

$z_1+z_2=1+\sqrt{3}i+\sqrt{3}+i\\=(1+\sqrt{3})+(1+ \sqrt{3})i\\|z_1+z_2|=\sqrt{(1+\sqrt{3})^2+(1+\sqrt{3})^2}\\=\sqrt{8+4\sqrt{3}}\\arg(1+\sqrt{3}+(1+ \sqrt{3})i)$

$tan^{-1}\left(\frac{1+ \sqrt{3}}{1+ \sqrt{3}}\right)= \frac{\pi}{4}\\z_1+z_1=2\sqrt{2+\sqrt{3}}cis\frac{\pi}{4}$

Glyde · Oct 16, 2015

InteGrand said:
$Given: $z_1 = 1+\sqrt{3}i$ and $z_2 = \sqrt{3}+i$.$

$$\therefore z_1 + z_2 = 1+\sqrt{3}i + \sqrt{3}+i = \left(1+\sqrt{3} \right)+i\left( 1 + \sqrt{3}\right)i$ (add the real and imaginary parts respectively to get the sum of two complex numbers)$

$So $z_1 + z_2 = \left(1+\sqrt{3} \right) (1+i)$. We have$

$$=\mathrm{arg} (z_1 + z_2 ) = \mathrm{arg}\left[ \left(1+\sqrt{3} \right) (1+i)\right]$$

$$=\mathrm{arg}(1+i)$ (the positive constant in front of $(1+i)$ does not affect the arg, since the angle is the same)$

$= \tan^{-1} \frac{1}{1} =\frac{\pi}{4}$

$, and $|z_1 + z_2|= \left | \left(1+\sqrt{3} \right) (1+i) \right |=\left |1+\sqrt{3} \right | \times |1+i| = \left(1+\sqrt{3} \right) \times \sqrt{1^2 + 1^2}= \left(1+\sqrt{3} \right)\sqrt{2}$.$

$Now that we now the modulus and argument of $z_1 + z_2$, we have $z_1 + z_2 = \left(1+\sqrt{3} \right)\sqrt{2} \left[\cos \frac{\pi}{4} + i\sin \frac{\pi}{4} \right] $.$

Thanks!! I just don't understand what you mean about the constant not affecting (1+i) ?

Glyde · Oct 16, 2015

Ambility said:
Hey! I just started complex numbers this week as well!

$z_1+z_2=1+\sqrt{3}i+\sqrt{3}+i\\=(1+\sqrt{3})+(1+ \sqrt{3})i\\|z_1+z_2|=\sqrt{(1+\sqrt{3})^2+(1+\sqrt{3})^2}\\=\sqrt{8+4\sqrt{3}}\\arg(1+\sqrt{3}+(1+ \sqrt{3})i)$

$tan^{-1}\left(\frac{1+ \sqrt{3}}{1+ \sqrt{3}}\right)= \frac{\pi}{4}\\z_1+z_1=2\sqrt{2+\sqrt{3}}cis\frac{\pi}{4}$

Yes that is what I have been getting however the answer isn't that, the last 2 posts have the right answer.

InteGrand · Oct 16, 2015

Glyde said:
Thanks!! I just don't understand what you mean about the constant not affecting (1+i) ?

$In general, if $\lambda$ is a real number and is positive, then for any complex number $z\neq 0$, the following is true: $\mathrm{arg}\left(\lambda z\right)=\mathrm{arg}(z)$ (if $z=0$, the argument is undefined). Geometrically, if $\lambda$ is a positive real number, the effect of multiplying $z$ by $\lambda$ is that we get a new complex number $\lambda z$ that points in the same direction as the original complex number $z$ (viewed as a vector from the origin), but has its length scaled by a factor of $\lambda$. Since it points in the same direction, the argument remains the same (if you haven't learnt the vector representation of complex numbers yet, don't worry, you'll learn it soon).$

InteGrand · Oct 16, 2015

Glyde said:
Ambility said:

Hey! I just started complex numbers this week as well! I might be wrong, but I'll give it a try.

$z_1+z_2=1+\sqrt{3}i+\sqrt{3}+i\\=(1+\sqrt{3})+(1+ \sqrt{3})i\\|z_1+z_2|=\sqrt{(1+\sqrt{3})^2+(1+\sqrt{3})^2}\\=\sqrt{8+4\sqrt{3}}\\arg(1+\sqrt{3}+(1+ \sqrt{3})i)$

$tan^{-1}\left(\frac{1+ \sqrt{3}}{1+ \sqrt{3}}\right)= \frac{\pi}{4}\\z_1+z_1=2\sqrt{2+\sqrt{3}}cis\frac{\pi}{4}$

Click to expand...

Yes that is what I have been getting however the answer isn't that, the last 2 posts have the right answer.

$The answer provided by Ambility and those provided by me and leehuan are equivalent; this is because your modulus can be simplified to our answers' one and vice versa (i.e. they're equal). You can try proving this as an exercise if you like.$

Glyde · Oct 16, 2015

InteGrand said:
$The answer provided by Ambility and those provided by me and leehuan are equivalent; this is because your modulus can be simplified to our answers' one and vice versa (i.e. they're equal). You can try proving this as an exercise if you like.$

Ohhh is that why you only did pythagorus on (1+i) becuase the positive number in front of it scales it's length ?
thanks for the help

Ambility · Oct 16, 2015

InteGrand said:
$The answer provided by Ambility and those provided by me and leehuan are equivalent; this is because your modulus can be simplified to our answers' one and vice versa (i.e. they're equal). You can try proving this as an exercise if you like.$

I have to ruin the fun.

$2\sqrt{2+\sqrt{3}}\\=\sqrt{8+4\sqrt{3}}=\sqrt{2}+ \sqrt{6}\\8+4\sqrt{3}=(\sqrt{2}+\sqrt{6})^2\\=2+2 \sqrt{12}+6\\=8+4\sqrt{3}$

I guess both moduli are equally correct? Is one more simple than the other?

InteGrand · Oct 16, 2015

Ambility said:
I have to ruin the fun.

$2\sqrt{2+\sqrt{3}}\\=\sqrt{8+4\sqrt{3}}=\sqrt{2}+ \sqrt{6}\\8+4\sqrt{3}=(\sqrt{2}+\sqrt{6})^2\\=2+2 \sqrt{12}+6\\=8+4\sqrt{3}$

I guess both moduli are equally correct? Is one more simple than the other?

$Probably $\sqrt{2}\left(1+\sqrt{3}\right)$ is simpler since it avoids a surd inside a surd, but it doesn't really matter. And yes, the moduli are equal, so both are correct.$

Glyde · Oct 16, 2015

Now I have to divide (z1+z2) by (z1-z2) I think it would be easier to complete this question using the more simplified mod-arg

kawaiipotato · Oct 16, 2015

Glyde said:
Now I have to divide (z1+z2) by (z1-z2) I think it would be easier to complete this question using the more simplified mod-arg

You should find the simplified expression for both z1+z1 and z1-z2, then multiply both numerator and denominator by the conjugate of the denominator.

Ambility · Oct 16, 2015

Glyde · Oct 16, 2015

kawaiipotato said:
You should find the simplified expression for both z1+z1 and z1-z2, then multiply both numerator and denominator by the conjugate of the denominator.

No wouldnt I just divide the modulis and take away the arguement in the denominator from the argument in the numerator
We learnt that shortcut today

kawaiipotato · Oct 16, 2015

Glyde said:
No wouldnt I just divide the modulis and take away the arguement in the denominator from the argument in the numerator
We learnt that shortcut today

Yes, you could do it that way if you were using polar form ( rcis(theta)). Preferably, I would just do what mentioned, because z1-z2 and z1+z2 have 'weird' real and imaginary parts, that probably wouldn't give a 'neat' value for their respective arguments when taking the inverse tan. (I haven't tried yet so their arguments might be simple)

edit: ah nvm I just looked at the above posts and the argument for z1+z2 is just pi/4 so z1-z2 would also be similar. Yeah, your way would be quicker

Glyde · Oct 16, 2015

kawaiipotato said:
Yes, you could do it that way if you were using polar form ( rcis(theta)). Preferably, I would just do what mentioned, because z1-z2 and z1+z2 have 'weird' real and imaginary parts, that probably wouldn't give a 'neat' value for their respective arguments when taking the inverse tan. (I haven't tried yet so their arguments might be simple)

edit: ah nvm I just looked at the above posts and the argument for z1+z2 is just pi/4 so z1-z2 would also be similar. Yeah, your way would be quicker

I just tried to do z1-z2 however the argument was 3pi/4 why is it this when I calculated the tan inverse to be -45!!! Shouldn't the answers argument be -pi/4!!!!????

InteGrand · Oct 16, 2015

Glyde said:
I just tried to do z1-z2 however the argument was 3pi/4 why is it this when I calculated the tan inverse to be -45!!! Shouldn't the answers argument be -pi/4!!!!????

$You need to add on $\pi$ to the inverse tan answer because the complex number is in the second quadrant. The rule for finding the principal argument $\theta$ of a complex number $z=x+iy$ ($x,y$ real numbers) is \textbf{not} actually just $\theta = \tan ^{-1}\left(\frac{y}{x}\right)$ in general; this rule is correct if and only if $x$ is positive. If $x$ is negative, then instead, you need to compute $ \tan ^{-1}\left(\frac{y}{x}\right)$ and then add or subtract $\pi$, depending on the sign of $y$ (if $y$ is positive, add $\pi$, and if $y$ is negative, subtract $\pi$). If $x$ is 0, then the argument is $\frac{\pi}{2}$ if $y>0$ and $-\frac{\pi}{2}$ if $y<0$. If $x$ and $y$ are both 0, then $z=0$, and the argument is undefined. (All these rules are described by the $\mathrm{atan2}$ function; see the link below. Basically, that is the complete rule for finding the (principal) argument of a general complex number $x+iy$ )$

https://en.wikipedia.org/wiki/Atan2#Definition_and_computation

$It is always true that $\tan (\mathrm{arg}(z)) = \frac{y}{x}$ (provided $x\neq 0$), but due to the fact that inverse tan is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, we need to make adjustments to get the argument (to get to the right quadrant) and can't just say $\theta = \tan ^{-1}\left(\frac{y}{x}\right)$ in general, so check the $\mathrm{atan2}$ page.$

kawaiipotato · Oct 16, 2015

Glyde said:
I just tried to do z1-z2 however the argument was 3pi/4 why is it this when I calculated the tan inverse to be -45!!! Shouldn't the answers argument be -pi/4!!!!????

The argument is 3pi/4.
Draw up the complex number z1-z2 on an argand diagram and label it z1-z2. You'll notice that it lies in the second quadrant. So the argument of it should be pi/2 (since coordinate axes are equal) + angle between the vector z1-z2 and the y axis which is arctan((sqrt3 - 1)(sqrt3 - 1)) = arctan(1)= pi/4. (I know the it the numerator should be negative but I'm considering only the lengths (positive value))
= pi/2 + pi/4 = 3pi/4

InteGrand · Oct 16, 2015

Complex number help! (2 Viewers)

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