Complex Numbers

chocolates-xed SG

got two questions

1. The complex number x + iy is such that (x+iy)^2 = i Find the possible vales of the real numbers x and y giving ur answers in exact form, Hence find the possible values of the complex number such that w^2 = -i

2.It is givn that -1 + 2i satisfies the cubic equation z^3 + 3z^2 + az +b =0 where a and b are real constants. find the value of a and show that b =5. Hence obtain all the exact values of the roots of the cubic equation.

qn 2 right, i know must use conjugate root theorem but i dunnno how to do this part

(z-1+2i)(z+1-2i) is it expand normally?

thanks:D

SBS261P SG

use factor theorem. there is a real root (z=-1) den after that factorise outi

wee_ws SG

Hi,

It is useful to remember that (z - c)(z - c*) = z^2 - 2 Re(c) z + cc*, where c is a complex root and c* is the conjugate of c. This concept (sum and product of roots) is actually covered in the new Additional Mathematics syllabus.

In Q2, c = -1 + 2i and c* = -1 - 2i, so our quadratic equation becomes z^2 + 2z + 5.

Now (z^2 + 2z + 5)(z + 5/b) = z^3 + 3z^2 + az + b.

Comparing the coefficients of z^2, 5/b + 2 = 3, so b = 5.

Thanks!

Cheers,
Wen Shih

blueftw SG

6. The complex number z is such that z=(a/1+i) + (b/1-3i) where a and b are real. If arg(z)= - pie/2 and modulus of Z =4 find the values of a and b, [a=2, b=-10)

9(i) state the value of e^(i(pie))

need help with these 2 qns

wee_ws SG

Hi,

For Q6, we first express z in the form of x + iy. x and y will be expressions involving a and b.

Next, we form two equations, since we have to solve for two unknowns, i.e.

tan (-pi/2) = y/x and x^2 + y^2 = 4^2.

For Q9, use the fact that e^(ix) = cos x + i sin x.

Try it, thanks!

Cheers,
Wen Shih

blueftw SG

kkays i got qn 9 already. the sin pi will = 0 then cos pi = -1 so e^i(pi) = -1

thanks