Question

NAME: Q1. (30pts) Solve the quadratic equation a+(1+4i)2-1- 13i 0 by realizing the following plan: (i) find the discriminant A of the equation: (i) write a program for a scientific calculator to obtain the polar form r(cos@+ i sin 6) of A and the first root As rcos-+isin of degree two of A: (ii) execute the program, fix the results, find another root A of A of degree two (before executing the program, make sure that your caleulator is set to Radian mode); (iv) find the roots of the quadratic equation.

Answer 1

Equation is of the form

azbzc0 where a 1, b 14i, c 1 13i

Then we have

b2-4ac (1+4i)2-4(1) (-1 13i) 11 60i

So we need to find (say)

-1160i x iy

This means

2y 11, ry 30 (xiy)2 =x2 - 2xyi -11602

So

30 2 900 y = = -11 2

Meaning
x 112900 0
so that
(x2 25) (r236) 0 25
as x and y must be real

So
5 y £6
meaning the two roots are
/-11 60i ±(5 + 6i)

Thus, the roots of our quadratic are

bt vb-4ac (14) (5+6i) 2a 2

That is, $\frac{6+10i}{2},\,\frac{-4-2i}{2}$

Which are

235i,-2-