CIVIL ENG MODULE TWO PAST PAPERS SECTION A: MATHEMATICS II Answer at least TWO questions from this section.

Panua $(1+j)^2$:

$$(1+j)^2 = 1^2 + 2(1)(j) + j^2 = 1 + 2j - 1 = 2j$$

Panua $(2-j)^2$:

$$(2-j)^2 = 2^2 - 2(2)(j) + j^2 = 4 - 4j - 1 = 3 - 4j$$

$$x(2j) + y(3-4j) = 3+10j$$ $$2xj + 3y - 4yj = 3+10j$$

Panga upande wa kushoto:
Sehemu halisi: $3y$
Sehemu ya kufikirika: $2xj - 4yj = (2x - 4y)j$
Hivyo mlinganyo unakuwa:

$$3y + (2x - 4y)j = 3+10j$$

Kulinganisha sehemu halisi:

$$3y = 3 \quad \text{(Mlinganyo 1)}$$

Kulinganisha sehemu za kufikirika:

$$2x - 4y = 10 \quad \text{(Mlinganyo 2)}$$

Kutoka Mlinganyo 1:

$$3y = 3$$ $$y = \frac{3}{3}$$ $$y = 1$$

Badilisha $y=1$ kwenye Mlinganyo 2:

$$2x - 4(1) = 10$$ $$2x - 4 = 10$$ $$2x = 10 + 4$$ $$2x = 14$$ $$x = \frac{14}{2}$$ $$x = 7$$

Thamani za $x$ na $y$ ni:

$$x=7, y=1$$