show that: The modulus and argument of the complex number $z_{1} = z^{2} - z,$ if $z = cos ϕ + i sin ϕ .$ is $2 | sin ϕ / 2 |, (\frac{3 π + 3 ϕ}{2})$ .

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Solution

$z = e^{i ϕ}$
Hence
$z^{2} - z$
$= e^{2 i ϕ} - e^{i ϕ}$
$= c o s 2 ϕ + i s i n 2 ϕ - c o s ϕ - i s i n ϕ$
$= - 2 s i n \frac{3 ϕ}{2} . s i n \frac{ϕ}{2} + i (2 s i n \frac{ϕ}{2} . c o s \frac{3 ϕ}{2})$
$= 2 s i n \frac{ϕ}{2} [- s i n \frac{3 ϕ}{2} + i c o s \frac{3 ϕ}{2}]$
$= 2 s i n \frac{ϕ}{2} [c o s (\frac{π}{2} + \frac{3 ϕ}{2}) + i s i n (\frac{π}{2} + \frac{3 ϕ}{2})]$ or $2 s i n \frac{ϕ}{2} [c o s (\frac{3 π}{2} + \frac{3 ϕ}{2}) + i s i n (\frac{3 π}{2} + \frac{3 ϕ}{2})]$ depending on the value of $ϕ$ .
Hence
$| z | = 2 | s i n \frac{ϕ}{2} |$
And $a r g (z) = \frac{3 π}{2} + \frac{3 ϕ}{2}$ or $\frac{π}{2} + \frac{ϕ}{2}$ depending on the value of $ϕ$ .

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show that: The modulus and argument of the complex number z1=z2−z, if z=cosϕ+isinϕ. is 2|sinϕ/2|,(3π+3ϕ2).

show that: The modulus and argument of the complex number $z_{1} = z^{2} - z,$ if $z = cos ϕ + i sin ϕ .$ is $2 | sin ϕ / 2 |, (\frac{3 π + 3 ϕ}{2})$ .