Question

The equation $\text{Im}(1-i)z=1$ represents the line

• A. $y=x+1$
• B. $y=1-x$
• C. $y=x-1$
• D. $y=x+2$
• E. $y=-x-1$

Hint:

Logic Tip: When working with geometric representations of complex equations, translating $z$ to $(x+iy)$ immediately turns a complex algebra problem into a familiar Cartesian geometry problem.

Answer 1

The Correct Option is A

Concept:
To analyze an equation in the complex plane, express the complex variable $z$ in its algebraic form $z = x + iy$, where $x$ is the real part and $y$ is the imaginary part. The operator $\text{Im}()$ extracts the imaginary part of the resulting complex expression.
Step 1: Substitute the algebraic form of z.
Let $z = x + iy$. We need to expand the expression $(1-i)z$: $$(1-i)z = (1-i)(x+iy)$$
Step 2: Expand the complex product.
Use standard algebraic distribution (FOIL) and remember that $i^2 = -1$: $$(1-i)(x+iy) = x + iy - ix - i^2y$$ $$(1-i)(x+iy) = x + iy - ix - (-1)y$$ $$(1-i)(x+iy) = x + iy - ix + y$$
Step 3: Separate into real and imaginary parts.
Group the terms with $i$ and without $i$: $$(1-i)z = (x + y) + i(y - x)$$ Here, the real part is $(x + y)$ and the imaginary part is $(y - x)$.
Step 4: Apply the condition given in the problem.
The problem states that the imaginary part of this expression equals 1: $$\text{Im}((1-i)z) = 1$$ $$y - x = 1$$ Rearranging for $y$ yields the equation of the line: $$y = x + 1$$