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Let z be a complex number such that z^5 = 1 and $$z \neq 1$$. Compute $$z + \frac{1}{z} + z^2 + \frac{1}{z^2}.$$

Let z be a complex number such that |z - 5 - i| = 5. Find the minimum value of $$|z - 1 + 2i|^2 + |z - 9 - 4i|^2.$$

Apr 19, 2019

#1
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$$z^5 = 1 \implies z = e^{\frac{2k\pi i}{5}}, k = 1,2,3,4\\ \dfrac{1}{z} = z^4\\ \dfrac{1}{z^2} = z^3\\ \text{Substitute }z = e^{\frac{2\pi i}5}\\ z+\dfrac1{z}+z^2+\dfrac1{z^2} = z+z^2+z^3+z^4 = \displaystyle \sum_{k=1}^4 e^{\frac{2k\pi i}{5}}$$

So the question is actually asking the sum of roots of the equation $$z^5-1 = 0$$, excluding 1.

Sum of roots of $$ax^5+bx^4+cx^3+dx^2+ex+f=0$$ is $$\dfrac{-b}{a}$$. But don't forget to exclude 1.

The required value = $$-\dfrac{0}{1} - 1$$ = -1.

Apr 20, 2019
#2
+7725
+1

Make use of the identity: $$z\cdot z^{*} = |z|^2$$.

$$(z-5-i)(z-5-i)^* = 25$$

Let z = x + y i.

$$((x-5)+(y-1)i)((x-5)+(1-y)i) = 25\\ (x-5)^2 + (y-1)^2 = 25$$

So the equation |z - 5 - i| = 5 actually represents a circle centered at z = 5 + i and with radius 5.

Expressing the given expression with x and y:

$$|z-1+2i|^2 + |z-9-4i|^2\\ =(x-1)^2+(y+2)^2 + (x-9)^2 + (y-4)^2\\ = x^2-2x+1+y^2+4y+4+x^2-18x+81+y^2-8y+16\\ =2x^2+2y^2-20x-4y+102\\ =2(x^2+y^2-10x-2y+102)\\ =2((x-5)^2+(y-1)^2+76)$$

And because $$(x-5)^2+(y-1)^2 = 25$$,

$$|z-1+2i|^2 + |z-9-4i|^2\\ =2(25+76)\\ =202$$

So the minimum value is 202.

Apr 20, 2019