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4. Let Zl and be two roots of the equation z2 + az + b = 0 , z being complex number. Further, assume that the origin, and form an equilateral triangle, then : (b) a 2 = 2b -3b (d) a2 = 4b — , then 5. If z and oare two non-zero complex numbers such that I I, and arg (z) —arg = zcois equal to . 6.
T( z2 )1/2, with the principal branch of the square root chosen, maps 0 to 0 and the hyperbola xy = 1 onto the hyperbola u2 ?v2 = 1. Solution. Let T(z) = Az + B be the desired linear transformation. Since T(0) = 0, we must have B = 0 . Next, let z = x+iy satisfy xy = 1 and w = u+iv satisfy u 2?v = 1.
Calculator Use. This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ? 0, using the quadratic formula. The calculator solution will show work using the quadratic formula to.
The function f is injective or one -to- one if every point in the image comes from exactly one elementinthedomain.Toshowafunctionisinjectiveprove x 1;x 2 2A and f„x 1 …
