Question 1.Firstly, let \(z_a=z-3-i\): (we will be using subscript \(a\) to avoid confusion)
Here's where you'll apply De Moivre's Theorem, which gives us:
\\[1em]\therefore r=3,\ \theta=\frac{2k\pi}{3})
Then, substitute the values for \(k\): (\(k=0,1,2\))
You should then get the following equations:
You should be familiar with converting to rectangular form, but if not, here's a refresher:
)
And since \(z_a=z-3-i\), we can now substitute the values we have found for \(z_a\) into the original equation, so that we will obtain solutions for \(z\) :
)=6+i\\<br />z_2=\left(-\frac{3}{2}+\frac{3\sqrt{3}}{2}i+(3+i)\right)=\frac{3}{2}+\frac{3\sqrt{3}+2}{2}i\\<br />z_3=\left(-\frac{3}{2}-\frac{3\sqrt{3}}{2}i+(3+i)\right)=\frac{3}{2}+\frac{2-3\sqrt{3}}{2}i)
Therefore, the solutions are as follows:
Question 2.\((|z|-1)(|z|-2)\leq0\) and \(Im(z)>0\)
\(Im(z)>0\) just signifies that the required region is to the right of the imaginary (\(y\)-axis) axis. This will be useful when sketching later.
\((|z|-1)(|z|-2)\leq0\) can be rewritten as \((\sqrt{x^2+y^2}-1)(\sqrt{x^2+y^2}-2)\leq0\)
From that, it can be derived that the two equations are circles of radius 1 and radius 2 respectively.
The required region would be between the \(x\)-values 1 and 2, as shown below.
(Apologies for the large image, I'm new here so I don't know how this formatting works.)
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