Solve for x: x2 – 5ix – 6 = 0.

We have, x2 – 5ix – 6 = 0Here, b2-4ac = (-5i)2-4×1×-6= 25i2+24 = -25+24 = -1Therefore, the solutions are given by Hence, x= 3i and x = 2i

Q15 of 202 Page 212

Solve for x: x² – 5ix – 6 = 0.

We have, x² – 5ix – 6 = 0

Here, b²-4ac = (-5i)²-4×1×-6

= 25i²+24 = -25+24 = -1

Therefore, the solutions are given by

Hence, x= 3i and x = 2i

Express in the form (a + ib).

Solve for x: (1 – i) x + (1 + i) y = 1 – 3i.

Find the conjugate of .

If z = (1 – i), find z^-1.