How IMUL instruction works in 8086 microprocessor?

Instruction is simmilar to the MUL except that operands are assumed to be signed numbers.

The source operand specified in the instuction is multiplied with accumulator.

Example:

IMUL BL

If AL=80H, BL=20H

content of AL is treated as signed no. Which is 2'compliment of 128

-128*32=-4096

2'compliment of it is stored in AX i.e. F000H