How do you multiply # (-7+3i)(1+3i) # in trigonometric form?

Question

How do you multiply #  (-7+3i)(1+3i) # in trigonometric form?

Emery Adkins · Accepted Answer

#(-7+ 3 i)(1+ 3 i) =24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i) #

Let #Z=a+i b ; Z=-7+ 3 i ; a=-7 ,b = 3# ;

#Z=-7+ 3 i# is in #2# nd quadrant.

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt((-7)^2+ 3^2)) =sqrt 58 #

# tan alpha =|b/a|= 3/7 or alpha =tan^-1(3/7) ~~ 0.4049#

#theta# is on #2# nd quadrant # :. theta=pi-0.4049#

# :. theta~~ 2.7367#. Argument , # theta ~~2.7367:. #

Trigonometrically expressed, it is

#r(cos theta+isintheta) = sqrt58(cos 2.74+i sin 2.74) #

#Z=1+ 3 i# is in #1# st quadrant.

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt(1^2+ 3^2)) =sqrt 10 #

# tan alpha =|b/a|= 3/1 or alpha =tan^-1(3) ~~ 1.249#

#theta# is on #1# st quadrant # :. theta=1.249#

Argument , # theta ~~1.249:. #

Trigonometrically expressed, it is

#r(cos theta+isintheta) = sqrt 10 (cos 1.25+i sin 1.25) #

#(-7+ 3 i)(1+ 3 i) = #

# sqrt58(cos 2.74+i sin 2.74) * sqrt 10 (cos 1.25+i sin 1.25)~~ #

#sqrt58 * sqrt 10 ( cos (2.74+1.25) + i sin(2.74+1.25) ~~#

#24.08 ( cos (3.99) + i sin(3.99) = (-16-18 i)# [Ans]

Delilah Aguilar · Answer

undefined To multiply $(-7+3i)(1+3i)$ in trigonometric form, we can first express each complex number in polar form.

For $-7+3i$, we have:
$$ r_1 = \sqrt{(-7)^2 + (3)^2} = \sqrt{49 + 9} = \sqrt{58} $$
$$ \theta_1 = \arctan\left(\frac{3}{-7}\right) = \arctan\left(-\frac{3}{7}\right) $$

For $1+3i$, we have:
$$ r_2 = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} $$
$$ \theta_2 = \arctan\left(\frac{3}{1}\right) = \arctan(3) $$

Then, we multiply the magnitudes and add the angles:
$$ r = r_1 \times r_2 = \sqrt{58} \times \sqrt{10} $$
$$ \theta = \theta_1 + \theta_2 = \arctan\left(-\frac{3}{7}\right) + \arctan(3) $$

Finally, we convert back to rectangular form using Euler's formula:
$$ z = re^{i\theta} $$

So, the product of $(-7+3i)(1+3i)$ in trigonometric form is $ z = \sqrt{580} e^{i(\arctan\left(-\frac{3}{7}\right) + \arctan(3))} $.