Question
Could someone explain #23 using De Moivre's theorem?
Uhons em can be föulnd m of a Powers and Roots Use De Moivre's Theorerm to express the trigonometric function Exercises 15 and 16 in terms of cos and sin . nctions in 15. cos4 17. Find the three cube roots of 1. 1 16. sin4 18. Find the two square roots of i. 19. Find the three cube roots of -8i. 20. Find the six sixth roots of 64. 21. Find the four solutions of the equation z4-272 + 4 = 0. 22. Find the six solutions of the equation z 6 + 2-3 + 2 = 0. 23. Find all solutions of the equation x4 + 4x2 16 0. 24. Solve the equation x4 + 1 = 0 Theory and Examples 25. Complex numbers and vectors in the plane Show with an Argand diagram that the law for adding complex numbers is the same as the parallelogram law for adding vectors. 26. Complex arithmetic with conjugates Show that th of the sum (product, or quotient) of two complex numbers, 2, is the same as the sum (product, or quotient) of conjugates of their ··Complex roots ofpolynonials with rea l coefficients come in complex-conjugate pairs a. Extend the results of Exercise 26 to show that if to show that f(z) = + alz + ao f(E) /(z) = anz" + an-lz"-! + is a polynomial with real coefficients ao, dn b. Ifz is a root ofthe equation f(z) 0, where/Chow that 2) is a polynomial with real coefficients as in part (a),set the conjugate z is also a root of the equation. ise the f(z) = 1 in part (a), show that tion. Hint: e fact Ise the fact
Explanation / Answer
x^4 +4x^2 +16 =0
let x^2 =p==>x^4 =p^2
p^2 +4p +16 =0
==>p =(-4 +sqrt(4^2 -4*1*16))/2,(-4 -sqrt(4^2 -4*1*16))/2
==>p =(-4 +sqrt(-48))/2,(-4 -sqrt(-48))/2
==>p =(-4 +i4sqrt(3))/2,(-4 -i4sqrt(3))/2
==>p =-2 +i2sqrt(3),-2 -i2sqrt(3)
==>p =2(-1 +isqrt(3)),2(-1 -isqrt(3))
==>p =4(-1/2 +i(sqrt(3))/2), 4((-1)/2 +i(-sqrt(3))/2)
==>p=4ei 2pi/3,4ei 4pi/3
==>x^2 =4ei 2pi/3, x^2=4ei 4pi/3
==>x=sqrt(4ei 2pi/3) ,x=sqrt(4ei 4pi/3)
==>x=2sqrt(ei 2pi/3) ,x=2sqrt(ei 4pi/3)
==>x=2(ei 2pi/(3*2)),2(ei 2pi/(3*2)+2pi/2),x=2(ei 4pi/(3*2)),2(ei 4pi/(3*2)+2pi/2)
==>x=2(ei pi/3),2(ei 4pi/3),x=2(ei 2pi/3),2(ei 5pi/3)
==>x = 1+i sqrt(3),x = -1-i sqrt(3),x = -1+i sqrt(3),x = 1-i sqrt(3)