Finding Argument of Complex Number Given Equations

NEILS BOHR
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Homework Statement


let z , w be complex nos. such that z + i ( conjugate of w ) = 0 and zw = pi . Then find arg z..


Homework Equations





The Attempt at a Solution


well i m unable to understand wat is meant by zw=pi...
 
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You sure you wrote the problem correctly? I found one very similar to this by searching via Google. The problem goes like this:
"Let z, w be complex numbers such that

\bar{z} + i\bar{w} = 0

and

arg(zw) = π.

Find arg(z)."
 
hmmm
well in the quesn it is given like this only...

but yeah without arg it doesn't make much of a sense...
 
It is possible for z*w = pi, but it's more likely that arg(z*w) = pi.
 
can u pleasez elaborate a little??
 
If z= r_ze^{i\theta_z} and w= r_we^{i\theta_w} then arg(zw)= \theta_z+ \theta_w= \piso \theta_z and \theta_w are supplementary angles.<br /> <br /> Saying that z+ i\overline{w}= 0 means that z= -i\overline{w} and so arg(z)= arg(i\overline{w})- \pi.<br /> <br /> Now, taking the conjugate of a complex number multiplies its argument by -1 and multiplying by i adds \pi/2 to the argument. That is, if w has argument \theta, then i\overline{w} has argument \pi/2- \theta. From the previous paragraph, arg(z)= arg(i\overline{w})+ \pi= \pi2- \theta+ \pi= -(\theta+ \pi/2). More than that, I don't believe you can say.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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