Question

2a. Solve the quadratic equation

3z² + (a - i)z + 3i = 0.

b. Solve the following system of equations of complex numbers:

z + iw = 1 + 2i

z - w = 1 -2i

Answer 1

a.

$$3z^2 + (a - i)z + 3i = 0.$$

$$D=(a-i)^2-4(3)(3i)=a^2-2ai-1-36i$$

$$=a^2-1-2(a+18)i$$

$$z_1=\dfrac{-a+i-\sqrt{a^2-1-2(a+18)i}}{6}$$

$$z_2=\dfrac{-1+i+\sqrt{a^2-1-2(a+18)i}}{6}$$

b.

$$z + iw = 1 + 2i$$ $$z - w = 1 -2i$$

$$w(i+1) =4i$$ $$z(1+i)= 3+3i$$

$$w=\dfrac{4i}{1+i}$$ $$z=3$$

$$w=2+2i$$ $$z=3$$