Real part of ((1+j)(e^jθ))=-1 solve for θ?

(1+j)(e^jθ)
Covert to rectangular form
(1+j)(cosθ+jsinθ)
={cosθ +j^2 sinθ} + j(sinθ+cosθ)
Just take the real part and set to -1 as mentioned in the problem
Cosθ - sinθ = -1
solve theta and you will get
θ=pi/2 or 90°

09228777220
Engr. Padilla