Find ##z## in the form ##a+bi## under Complex Numbers

chwala
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Homework Statement
see attached
Relevant Equations
Complex numbers
1646184520481.png


For part (a),
##z##=##\dfrac {3+i}{3-i}## ⋅##\dfrac {3+i}{3+i}##
##z##=##\dfrac {4}{5}##+##\dfrac {3}{5}i##

part (b) no problem as long as one understands the argand plane...

For part (c)
Modulus of ##z=1##
and Modulus of ##z-z^*##=##\frac{6}{5}i##
 
Last edited:
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The modulus is a real number, so it can't be ##\frac 6 5 i##.
 
PeroK said:
The modulus is a real number, so it can't be ##\frac 6 5 i##.
True, its supposed to be ##\dfrac {6}{5}##
 
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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