Complex algebra?

(a)if w=4+3i and z= -3+2i,express


each of the following complex


numbers in cartesian form:-


(i)wz


(ii)conjugate of z


(iii)z^(-1)


(iv)w/z


(b)find a polynomial with real


coefficients whose roots are


(-2+7i) and (-2-7i)


(c)express the complex number


(-64) in polar form

Complex algebra?
Question a)


(i) wz = (4 + 3i) (-3 + 2i) = -12 - i - 6 = -18 - i





(ii) conj z = -3 - 2i





(iii) z^(-1) = 1/(-3 + 2i) = (-3 - 2i)/(-3 + 2i)(-3 - 2i)


= (-3 - 2i)/(9 + 4)


= (-1/13)(3 + 2i)





(iv) w/z = (4+3i)(-3 - 2i) / (-3 + 2i)(-3 - 2i)


= ( -12 -17i + 6) /13 = (- 6 -17i)/13 = -1/13(6 + 17i)





Question b)


Let z1 = - 2 + 7i and z2 = - 2 - 7i


(z - z1)(z - z2) = 0


z² - (z1 + z2) z + z1z2 = 0


z² - (-4)z + (-2 + 7i)(-2 -7i) = 0


z² + 4z + (4 + 49) = 0


z² + 4z + 53 = 0





Question c)


- 64 = 64(cos180° - i sin180°) = 64 at 180°
Reply:Teamwork Richard!!!!!!


Cheers! Report It

Reply:i) (4 + 3i)(-3 + 2i) = -12 + 8i - 9i + 6i² = 12 - i - 6 = 6 - i





ii) -3 - 2i





iii) 1/(-3 + 2i) × (-3 - 2i)/(-3 - 2i) = (-3 - 2i)/(9 - 4i²) = (-3 - 2i)/(9 + 4) = (-3 - 2i)/13





iv) (4 + 3i)/(-3 + 2i) × (-3 - 2i)/(-3 - 2i) = (-12 - 8i - 9i - 6i²)/(9 - 4i²) = (-12 - 17i + 6)/(9 + 4) = (-6 - 17i)/13





b)





(x - (-2 + 7i))(x - (-2 - 7i)) = x² - (-2 - 7i)x - (-2 + 7i)x + (-2 + 7i)(-2 - 7i)) = x² + 2x + 7ix + 2x - 7ix + 4 + 14i - 14i - 49i² = x² + 4x + 4 + 49 = x² + 4x + 53





c) you sure you meant -64? In polar form that would just be r = 64, theta = π...did you forget the imaginary part?
Reply:(4+3i)(-3+2i) = -12 +8i -9i +6i^2 = -18-i





cnj = -3-2i





1/z = 1/( -3+2i) = (-3-2i)/ 5 multiply num and den.by the conj of z





(4+3i)/(-3+2i) = (4+3i)( -3-2i) /5 = 1/5 ( (-6-17i)





(x-(-2-7i))( x- (-2+7i))= ((x+2)+7i)*((x+2) -7i) = (x+2)^2 +49=


x^2+4x+53





(-64) in polar has mod 64 and angl 180 deg.