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A product of 2 complex number will be a real number if and only if they are conjugates.

From $xy+yz+zx=24$ , the sum of all the products of two complex numbers is a real number. Let $y$ be a conjugate of $x$ , $y=x^*$ and $x=a+ib$ where $a$ and $b$ are real numbers. $xy+yz+zx=24$ $xx^* +z(x+x^*)=24$ $(a^2+b^2)+z(a+ib+a-ib)$ $a^2+b^2+ z(2a)=24...(3)$ Since $z(2a)$ must be a real number, $z$ must be a real number too. From $|x|=|y|=|z|$ , $|x|=|x^*|=|z|$ $\sqrt{a^2+b^2}=z$ $z^2=a^2+b^2 ...(1)$ From $xyz=64$ , $xx^*z=64$ $z(a^2+b^2)=64 ...(2)$ Substituting (1) into (2), $z^3=64$ $z=4$ From (1), $a^2+b^2=4^2$ Substitute $z$ and $a^2+b^2$ into (3), $16+4(2a)=24$ $2a=2$ Thus, $|x+y+z|=a+ib+a-ib+z=2a+z=6$

We use $|xyz| = |x| |y| |z| = |x|^3 = 64 \Rightarrow |x| = 4 = |y|=|z|$

So, we now have: $|x|^2 = x \overline{x} = 16 \Rightarrow \frac{1}{x} = \frac{\overline{x}}{16}$ . Same for complex numbers $y$ & $z$ .

Therefore, $xy + yz + zx = xyz (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \Rightarrow 64 (\frac{\overline{x}+\overline{y}+\overline{z}}{16}) = 24$

$\Rightarrow \overline{x}+\overline{y}+\overline{z} = 6 = \overline{x+y+z}$

$\Rightarrow |x+y+z|=|\overline{x+y+z}| = \boxed{6}$ .

For a complex number $w$ ,

$\frac{1}{w} = \frac{\bar{w}}{w \bar{w}} = \frac{\bar{w}}{|w|^2}$

Let $|x| = |y| = |z| = a$

$|xyz| = |64| = 64 \Rightarrow a = 4$

$xy + yz + zx = xyz(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = xyz(\frac{\bar{x} + \bar{y} + \bar{z}}{16}) = 24$

$\Rightarrow \bar{x} + \bar{y} + \bar{z} = 6 \Rightarrow |x+y+z| = |\bar{x} + \bar{y}+\bar{z}| = \fbox{6}$

We know that all cubics have one real root and two complex roots which are complex conjugates of each other. That being said, letting $x=a+bi$ and $y=a-bi$ , $xy=a^2+b^2=|x|^2=z^2$ . Hence, $xyz=z^3=64\Rightarrow z=4$ . By the third condition, we have $xy+yz+zx=z^2+z(x+y)=z(x+y+z)=4(x+y+z)=24$ , hence $|x+y+z|=x+y+z=\frac{24}{4}=\boxed{6}$ .

$|x|=|y|=|z| \Rightarrow |x|^2=|y|^2=|z|^2 \Rightarrow x\overline{x}=y\overline{y}=z\overline{z}=k(say)$

$xyz=64 \Rightarrow \overline{x}\overline{y}\overline{z}=64$

Now, $k^3=(x\overline{x})(y\overline{y})(z\overline{z})=64\times 64 \Rightarrow k=16$

Again, $\frac{xy+yz+zx}{xyz}=\frac{24}{64} \\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{8}\\ \Rightarrow \frac{\overline{x}+\overline{y}+\overline{z}}{k}=\frac{3}{8} \\ \Rightarrow \overline{x+y+z}=\frac{3k}{8}=\frac{3\times 16}{8}=6 \\ \Rightarrow |x+y+z|=6$

let $x=r_{1}\exp{i\theta_{1}}$ , $y=r_{2}\exp{i\theta_{2}}$ , $z=r_{3}\exp{i\theta_{3}}$ .... hence from first condition $r_{1}=r_{2}=r_{3}$ , now by 2nd $r^{3}(\exp{(\theta_{1}+\theta_{2}+\theta_{3})})=64......1$ , from here by comparing real and imaginary part one can get $r=4$ by 3rd condition, $r^{2}(\exp{i(\theta_{1}+\theta_{2})}+\exp{i(\theta_{2}+\theta_{3})}+\exp{i(\theta_{3}+\theta_{1})} )=24........2$ now equation(2/1): $\exp{-(i\theta_{3})}+\exp{-(i\theta_{1})}+\exp{-(i\theta_{1})}=\frac{24r}{64}=\frac{3}{2}$ comparing real and imaginary part of above equation: $\cos(\theta_{1})+\cos(\theta_{2})+\cos(\theta_{3})=\frac{3}{2}$ and $\sin(\theta_{1})+\sin(\theta_{2})+\sin(\theta_{3})=0$ hence: $x+y+z =r\Big((\cos(\theta_{1})+\cos(\theta_{2})+\cos(\theta_{3})) +i (\sin(\theta_{1})+\sin(\theta_{2})+\sin(\theta_{3}))\Big)$ so the ans is $= \frac{4\times3}{2}=\boxed{6}$

After some contemplation, we notice that since $xyz=64$ , either all of $x,y,z$ are purely real (i.e. $Im(x)=Im(y)=Im(z)=0$ ) or one of them is purely real and two are complex-conjugates (i.e. WLOG $Im(x)=-Im(y) \ne 0$ and $Im(z) = 0$ ).

Note: For a complex number $w$ , here we use purely real to denote $Im(w)=0$ and purely complex to denote $Im(w) \ne 0$ .

*Case 1: Each of $x,y,z$ is *purely real **

We have that $|x|=|y|=|z|$ . Since in this case the imaginary part is 0, we are dealing with plain old real numbers. This must mean that $x=y=z$ . We are also given that $xyz=64 \implies x=y=z=4$ . Plugging these values in to the third given equation we find: $xy+yz+zx=3(4^2)=48 \ne 24$ . This is a contradiction and so we move on to our last case.

*Case 2: Two of the numbers are complex-conjugates while the other is *purely real **

To give you a better understanding as to why this must be the only other case ( Proof ): Since we have $xyz=64=64+(0)i$ , and not all of $x,y,z$ are purely real , in order for the product of 3 complex numbers to turn out to be purely real , $Im(xyz)=0$ . If we have that each of $x,y,z$ has a non-zero imaginary part, then the imaginary part of the product will also be non-zero (this can be easily confirmed through multiplication and checking the coefficient of $i$ ). If one of $x,y,z$ is purely complex and two are purely real , then once again we (obviously) have that $Im(xyz) \ne 0$ . From Case 1 , we already know that if none of $x,y,z$ is purely complex then the given conditions are not satisfied. Hence the only case we are left with is if two of $x,y,z$ are purely complex while the other is purely real . Now in this situation, the only way $Im(xyz)=0$ is if the two purely complex numbers (WLOG, assume that $x$ and $y$ are purely complex ), when multiplied, have that $Im(xy)=0$ . This is only going to happen if $lm(x)=-lm(y) \implies y = \overline{x}$ because otherwise, we will once again have that $Im(xy) \ne 0$ (This too, can be confirmed by letting $x$ and $y$ be complex numbers and then multiplying them and checking $Im(xy)$ ).

Now let $x=a+bi$ , $y=\overline{x}=a-bi$ and $z=c \ | c \in \mathbb{R}$ . But we know that $|x|=|y|=\sqrt{a^2+b^2}=|z| \implies c = \sqrt{a^2+b^2}$ . Then:

$xyz=c(a+bi)(a-bi)=\sqrt{a^2+b^2}(a^2+b^2)=64$ . Now let $u=\sqrt{a^2+b^2}$ , then we have:

$u^3=64 \implies u = \sqrt{a^2+b^2}=4$ .

Considering the last given equation, we have:

$xy+yz+xz=xy+z(x+y)=(a+bi)(a-bi)+\sqrt{a^2+b^2}(a+bi + a - bi)=a^2+b^2+2a \sqrt{a^2+b^2} = 24$

We know that $\sqrt{a^2+b^2} = 4 \implies a^2+b^2=16$ . Making these substitutions yields:

$\implies (16)+2a(4)=16+8a=24 \implies a = 1$ . Now we are asked to find $|x+y+z|=|(a+bi)+(a-bi)+ \sqrt{a^2+b^2}|=|2a+\sqrt{a^2+b^2}|=2a+\sqrt{a^2+b^2}$ . Substituting for $a$ and $\sqrt{a^2+b^2}$ gives us:

$\implies |x+y+z|=2(1)+4=\boxed{6}$ . And we are done.

From (1) and (2), we get that $|x| = |y| = |z| = 4$

Since the product is real, I suspected that this was composed of a pair of complex conjugates and a real number. Thus, suppose $x = 4$ . Then $yz = 16$ and $4(y+z) + 16 = 24$ $4(y+z) = 8$ $y+z = 2$

We must show that such a $y$ and $z$ exist before using the above result.

Since $yz \in \mathbb{R}$ , we choose $y$ and $z$ as complex conjugates whose sum is $2$ : $y = 1 + \sqrt{15}i$ $z = 1 - \sqrt{15}i$

Therefore $y + z = 2$ and $x + y + z = \boxed{6}$

let the three complex no. be:

x= r e^{i a1}, y=re^{i a2}, z=re{i a3} (because |x|=|y|=|z|)

now, xyz= 64 i.e r^{3} e^{i(a1+a2+a3)}=64

this implies, r^{3}=64, i.e r=4 & sin(a1+a2+a3)=0

hence, a1+a2+a3=0 -..1

xy+yz+xz=24 i.e r^{2}( cos(a1+a2)+cos(a2+a3)+cos(a1+a3) +i(sin(a1+a2)+sin(a2+a3)+sin(a3+a1)) )=24

this implies, ( cos(a1+a2)+cos(a2+a3)+cos(a1+a3) +i(sin(a1+a2)+sin(a2+a3)+sin(a3+a1)) )=3/2

this implies, cos(a1+a2)+cos(a2+a3)+cos(a1+a3)=3/2 & sin(a1+a2)+sin(a2+a3)+sin(a3+a1)=0

Applying result 1 in above 2 eq,

Cos(a1)+cos(a2)+cos(a3)=3/2

& sina1+sina2+sina3=0

We have to find |x+y+z|, which is nothing but,

|r( (cosa1+cosa2+cosa3) + i(sina1+sina2+sina3))|=|4*3/2|

= 6

As given,

IxI = IyI = IzI, ......................(i)

xyz = 64 , .......................(ii)

xy + yz + xz = 24. .........................(iii)

We have to find Ix + y + zI = k(say).

From (i), we can see that their modular value is same and from (ii), it is observed that one parameter let x is real and other two y & z are complex conjugates.

Then, x * (IyI^2) = 64,

x * (IxI^2) = 64 .............from(i)

Now we can say that x = 4 and IyI = IzI = 4. .................(iv)

Now from (iii), we have xy + yz + xz = 24

x(y+z) + IyI^2 = 24

4 * (2Re(y)) = 8

Re(y) = 1

So, Re(z) = 1, Im(y) = sqrt(15)i , Im(z) = -sqrt(15)i

Therefore, x = 4, y = 1 + sqrt(15)i, z = 1 - sqrt(15)i.

So, k = 6 (Ans.)

Here is how I solved it: First, notice that all the norms of x,y,z are 4, so we can scale everything down by 4, and have the equation .|X| |Y||Z|=1 xy+yz+xz=3/2 The last equation becomes 1/x+1/y+1/2=3/2. Since x,y,z are on the unit circle 1/x=x,1/y=y,1/z=z, , we have x|+Y|+Z|=3/2, take conjugate on both sides, we get x+Y+Z=3/2 , scale everything back by 4, we get the final answer is .6

The first thing to notice is that we are concerned with the absolute value of the complex numbers. Therefore, rather than writing them in the usual form ( $a + bi$ ), it will be helpful to write them in polar form, as it will give us cleaner products. So, we define:

$x = r(\cos\alpha + i\sin\alpha)$

$y = r(\cos\beta + i\sin\beta)$

$z = r(\cos\gamma + i\sin\gamma)$

$xyz = r^3[\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)] = 64$

As $64$ has no complex part, $\sin(\alpha + \beta + \gamma) = 0 \implies \alpha + \beta + \gamma = 0$ This means, $\cos(\alpha + \beta + \gamma) = 1$ . Therefore, $r = 4$ .

Now, we expand out the second equation to obtain:

$xy + yz + zx = r^2[\cos(\alpha + \beta) + \cos(\beta + \gamma) + \cos(\gamma + \alpha)] + r^2i[\sin(\alpha + \beta) + \sin(\gamma + \alpha) + \sin(\beta + \gamma)] = 24$

As $\alpha + \beta + \gamma = 0$ , we make the following substitutions:

$\alpha + \beta = -\gamma$

$\beta + \gamma = -\alpha$

$\gamma + \alpha = -\beta$

$r^2 = 16$

Upon substituting and simplifying, we get:

$16(\cos\alpha + \cos\beta + \cos\gamma) -16i(\sin\alpha + \sin\beta + \sin\gamma) = 24$

As $24$ has no imaginary part,

\sin\alpha + \sin\beta + \sin\gamma = 0\tag1

We can simplify the rest of the equation and get,

\cos\alpha + \cos\beta + \cos\gamma = 1.5 \tag2

Multiplying $(1)$ by $i$ and adding to $2$ , we get:

$\cos\alpha + \cos\beta + \cos\gamma + i(\sin\alpha + \sin\beta + \sin\gamma) = 1.5$

Multiplying both sides by $r = 4$ , and manipulating gives:

$r(\cos\alpha + i\sin\alpha) + r(\cos\beta + i\sin\beta) + r(\cos\gamma + i\sin\gamma) = x + y + z = 6$

$|6| = \boxed{6}$ .

Let |x|= |y|= |z| = r. Also let x = re^iα, y = re^iβ and z = re^iγ. Given xyz = 64 gives 64 = re^i(α+β+γ) or r = 4 ------------- (1) and (α+β+γ) = 2π-------------------------------(2) Also it is given that xy+yz+xz= 24; which gives using (1), 16 {e^i(α+β) +e^i(β+γ) +e^i(α+γ)} = 24 or {e^i(α+β) +e^i(β+γ) +e^i(α+γ)} = (3/2) or, equating real and imaginary parts we get cos(α+β) + cos(β+γ)+cos(α+γ) = 3/2 ---------------------- (3), and sin(α+β) + sin(β+γ)+sin(α+γ) = 0 -----------------------------(4) or using(2), (3) becomes) cos{(2π)-γ} +cos{(2π)-α} +cos{(2π)-β} = 3/2 ------------ (5) or cos α +cos β + cos γ= 3/2 and similarly (4) becomes sin α +sin β + sin γ= 0 ----------------------------------------... (6) so x+y+z= 4 {(cos α +cos β + cos γ)+i (sin α +sin β + sin γ)} or using (5) and (6) we get x+y+z = 4 {(3/2) + i*0} or x+y+z = 6 or |x+y+z| = 6

Firstly note that guys, the first job is to get out as much information as possible out of the given. $x,y,z$ are complex nos, and further $|x|=|y|=|z|$ , so taking modulus on both sides of the second condition gives $|xyz|=|64|$ .......and using the fact that for complex nos $a,b$ taking modulus $|ab|=|a|\times|b|$ gives $|x|\times|y|\times|z|=64$ , using $|x|=|y|=|z|$ , gives $|x|^{3} =64, |x|=4$ .

So first condition gives $|x|=|y|=|z|=4$ . Now the job is to find $|x+y+z|$ . Call it $t$ , a positive real number. So $|x+y+z|=t$ . Squaring both sides and using the fact that for a complex number $z$ , $|z|^{2}=z\times \bar z$ , so $|x+y+z|^{2}=t^{2}$ , where $\bar z$ denotes complex conjugate, giving $(x+y+z)(\bar x+\bar y+\bar z)=t^{2}...............2$ . Expanding it and using the facts that $x \bar x=4^{2}=16$ gives $16+16+16+(x\bar y+x\bar z+y\bar x+y\bar z+z\bar y+z\bar x)=t^{2}$ . Now the problem reduces down to finding value of $(x\bar y+x\bar z+y\bar x+y\bar z+z\bar y+z\bar x)=p(say)......................1$ Now look, $p$ has to be found from the given conditions here lies the essence. Taking the conjugate on both sides of the third equation gives $\bar x\bar y+\bar y\bar z+\bar z\bar x=24......................a$ . Call the 3rd one b. $xy+yz+zx=24..................b$ .Multiplying a and b, note that terms like $xy \times \bar x\bar y =(|x||y|)^{2}=256$ and terms like $xy \times \bar z\bar y =|y|^{2} \bar x\bar z=16\bar x\bar z$ . So writing the entire thing after doing all multiplication and using the above results for other terms its got that $3\times 256+16p=576, p= -12$ . This step requires only algebraic manipulation. Plugging this into 2 gives $48-12=t^{2},t=6$ since $t$ is positive. So the answer is $6$ .