Question

Consider w=sqrrt2/2(cos(225°) + isin(225°)) and z = 1(cos(60°) + isin(60°)). What is w+ z expressed in rectangular form?

Answer 1

Answer:

C

Step-by-step explanation:

Answer 2

Answer:

Option (3)

Step-by-step explanation:

w = $\frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]$

Since, cos(225) = cos(180 + 45)

                          = -cos(45) [Since, cos(180 + θ) = -cosθ]

                          = -$\frac{\sqrt{2}}{2}$

sin(225) = sin(180 + 45)

             = -sin(45)

             = -$\frac{\sqrt{2}}{2}$

Therefore, w = $\frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]$

                      = $-\frac{2}{4}(1+i)$

                      = $-\frac{1}{2}(1+i)$

z = 1[cos(60) + i(sin(60)]

  = $[\frac{1}{2}+i(\frac{\sqrt{3}}{2})$

  = $\frac{1}{2}(1+i\sqrt{3})$

Now (w + z) = $-\frac{1}{2}(1+i)+\frac{1}{2}(1+i\sqrt{3})$

                   = $-\frac{1}{2}-\frac{i}{2}+\frac{1}{2}+i\frac{\sqrt{3}}{2}$

                   = $\frac{(i\sqrt{3}-i)}{2}$

                   = $\frac{(\sqrt{3}-1)i}{2}$

Therefore, Option (3) will be the correct option.