How do you write the following in trigonometric form and perform the operation given #(sqrt3+i)(1+i)#?

Question

Charles Autterson · Accepted Answer

# (\sqrt{3}+i)(1+i)=2\ text{cis}(\pi/6) \sqrt{2} \ text(cis)(pi/4)#

#= 2 \sqrt{2}\ text{cis} (pi/6 + pi/4)= 2 \sqrt 2\ text{cis}({5pi}/12)# I don't really get these questions. These are easy numbers to multiply in rectangular form -- why make it harder ? The two factors are the two biggest cliches in trig, 30/60/90 and 45/45/90, so it's OK to just write down the trigonometric forms and multiply without too much thought, as I did above. Let's turn each factor to trigonometric form, which is #z = r(cos theta + i sin theta) = r\ text{cis}\ theta,# basically polar coordinates, but still a rectangular form. Let's remember how to turn a number into trigonometric or polar form: # a + bi = r(cos theta + i sin theta)# We want to solve #a=rcos theta#, #b=r sin theta#. We get # r = sqrt{a^2 + b^2}# #theta = text{arctan2}(b //, a)# Usually folks write a plain arctan for the angle, but I want to make clear we need to two parameter, four quadrant inverse tangent to properly convert to polar. The odd #//,# notation reminds us it's two parameters and which is which. It's not much of an issue in this question as both factors are in the first quadrant, positive sine and cosine. #|sqrt{3}+i|=\sqrt{3+1}=2, angle (sqrt{3]+i) =text{arctan2}(1//,3) =pi/6# #|1+i|=\sqrt{1+1}=sqrt{2}, angle(1+i)=text{arctan2} (1//,1) = pi/4# # (\sqrt{3}+i)(1+i) = ( 2 \ text{cis} (pi/6) )(sqrt{2} \ text{cis} (\pi/4) )# # = 2 sqrt{2} \ text{cis} (pi/6+\pi/4)# # = 2 sqrt{2} \ text{cis}({5pi}/12) # We'll leave it there, but we know that equals #(\sqrt{3}+i)(1+i)=(\sqrt{3}-1) + i(\sqrt{3} + 1)#

Delilah Aguilar · Answer

undefined To write the expression $ (\sqrt{3} + i)(1 + i) $ in trigonometric form and perform the operation, we first multiply the complex numbers together:

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

Next, we simplify the expression by combining like terms and using the fact that $ i^2 = -1 $:

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

Now, to express this complex number in trigonometric form, we need to find its magnitude (or modulus) and argument (or angle). The magnitude can be calculated using the Pythagorean theorem:

$$
|z| = \sqrt{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2}
$$

$$
|z| = \sqrt{3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1}
$$

$$
|z| = \sqrt{8} = 2\sqrt{2}
$$

To find the argument, we use the arctan function:

$$
\theta = \arctan\left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right)
$$

$$
\theta = \arctan\left(\frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\right)
$$

$$
\theta = \arctan\left(\frac{4 + 2\sqrt{3}}{2}\right)
$$

$$
\theta = \arctan\left(2 + \sqrt{3}\right)
$$

Finally, we express the complex number in trigonometric form:

$$
z = 2\sqrt{2} \left(\cos(\arctan(2 + \sqrt{3})) + i \sin(\arctan(2 + \sqrt{3}))\right)
$$