Complex Number Questions (1 Viewer)

MC Squidge · Mar 12, 2009

i) If z is a complex number such that z=k(cos@+isin@), where k is real, show that arg(z+k)=@/2

ii) Deduce that if z1, z2 and z3 are ANY three complex numbers at the vertices of an equilateral triangle then

(z1)^2+(z2)^2+(z3)^2=z1z2+z2z3+z1z3

iii) Given that 1+z+z^2+z^3=(1-z^4)/(1-z) and that z=cos@+isin@

prove 1+cos@+cos2@+cos3@=1/2(1+(sin(7@/2)/sin(@/2)))

hint 2sinAsinB=cons(A-B)-cos(A+B)

Trebla · Mar 13, 2009

iii)
$1+z+z^2+z^3=1+cis\,\theta+cis\,2\theta+cis\,3\theta\\=(1+\cos\theta+\cos2\theta+\cos3\theta)+i(\sin\theta+\sin2\theta+\sin3\theta)\\$Also, $\\1+z+z^2+z^3 = \dfrac{z^4-1}{z-1}\\=\dfrac{\cos4\theta+i\sin4\theta - 1}{\cos\theta+i\sin\theta-1}\\=\dfrac{1-2\sin^22\theta+2i\sin2\theta\cos2\theta-1}{1-2\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}-1}\\= \dfrac{\sin2\theta(\sin2\theta-i\cos2\theta)}{\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}\\=\dfrac{\sin2\theta(\cos(\frac{\pi}{2}-2\theta)-i\sin(\frac{\pi}{2}-2\theta))}{\sin\frac{\theta}{2}(\cos(\frac{\pi}{2}-\frac{\theta}{2})-i\sin(\frac{\pi}{2}-\frac{\theta}{2}))}\\=\dfrac{\sin2\theta(\cos(-\frac{\pi}{2}+2\theta)+i\sin(-\frac{\pi}{2}+2\theta))}{\sin\frac{\theta}{2}(\cos(-\frac{\pi}{2}+\frac{\theta}{2})+i\sin(-\frac{\pi}{2}+\frac{\theta}{2}))}\\$

$=\dfrac{\sin2\theta}{\sin\frac{\theta}{2}}\left (\cos\left(-\frac{\pi}{2}+2\theta+\frac{\pi}{2}-\frac{\theta}{2}\right)+i\sin\left(-\frac{\pi}{2}+2\theta+\frac{\pi}{2}-\frac{\theta}{2}\right)\right)\\ =\dfrac{\sin2\theta}{\sin\frac{\theta}{2}} \left (\cos\frac{3\theta}{2}+i\sin\frac{3\theta}{2} \right )\\$Hence$\\(1+\cos\theta+\cos2\theta+cos3\theta)+i(\sin\theta+\sin2\theta+\sin3\theta)=\dfrac{\sin2\theta}{\sin\frac{\theta}{2}} \left (\cos\frac{3\theta}{2}+i\sin\frac{3\theta}{2} \right )\\$Equate real parts$\\1+\cos\theta+\cos2\theta+\cos3\theta=\dfrac{\sin2\theta\cos\frac{3\theta}{2}}{\sin\frac{\theta}{2}}\\=\dfrac{\sin(2\theta+\frac{3\theta}{2})+\sin(2\theta-\frac{3\theta}{2})}{2\sin\frac{\theta}{2}}\\=\dfrac{\sin\frac{7\theta}{2}+\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\\=\dfrac{1}{2}\left(1+\dfrac{\sin\frac{7\theta}{2}}{\sin\frac{\theta}{2}}\right)$

gurmies · Mar 13, 2009

(i) This is quite easily done geometrically. If you check out my diagram, you notice that since |z| = k, and arg (z+k) from origin is also k units, you can form an isosceles triangle. Therfore, 2 arg (z+ k) = @ [Exterior angle of a triangle]. Therfore arg (z+k) = @/2

Complex Number Questions (1 Viewer)

MC Squidge

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