# Find (Z1)(Z2) Leave answer in polar form

• Jul 30th 2008, 09:53 AM
Cyberman86
Find (Z1)(Z2) Leave answer in polar form
Given:
Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

Find (Z1)(Z2) Leave answer in polar form

This is what i got so far and i don't know if i started right.

=sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
=sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))

and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.
• Jul 30th 2008, 10:15 AM
TheEmptySet
Quote:

Originally Posted by Cyberman86
Given:
Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

Find (Z1)(Z2) Leave answer in polar form

This is what i got so far and i don't know if i started right.

=sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
=sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))

and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.

Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))

Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))

$z_1=\sqrt{6}e^{\frac{5 \pi}{6}i}$

$z_2=\sqrt{2}e^{\frac{5 \pi}{4}i}$

So we get

$z_1 \cdot z_2=\sqrt{6}e^{\frac{5 \pi}{6}i} \cdot \sqrt{2}e^{\frac{5 \pi}{4}i}=2\sqrt{3}e^{\frac{25 \pi}{12}i}=2\sqrt{3}e^{\frac{\pi}{12}i}=2\sqrt{3}\ left[ \cos\left( \frac{\pi}{12}\right)+i\sin\left( \frac{\pi}{12}\right)\right]$
• Jul 30th 2008, 10:17 AM
red_dog
If $z_1=r_1(\cos\phi_1+i\sin\phi_1)$ and $z_1=r_2(\cos\phi_2+i\sin\phi_2)$ then

$z_1z_2=r_1r_2[\cos(\phi_1+\phi_2)+i\sin(\phi_1+\phi_2)]$
• Jul 30th 2008, 10:50 AM
Soroban
Hello, Cyberman86!

Quote:

Given: . $\begin{array}{ccc}z_1&=& \sqrt{6}\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) \\ \\[-3mm]z_2 &=& \sqrt{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right) \end{array}$

Find: $(z_1)(z_2)$. .Leave answer in polar form.

We are expected to know this theorem . . .

If . $\begin{array}{ccc}z_1 &=& r_1\left(\cos\theta_1 + i\sin\theta_2\right) \\ z_2 &=& r_2\left(\cos\theta_2 + i\sin\theta_2\right) \end{array}$
. . . then: . $(z_1)(z_2) \;=\;(r_1\cdot r_2)\bigg[\cos\left(\theta_1+\theta_2\right) + i\sin\left(\theta_1+\theta_2\right)\bigg]$
. . . . . . . . . . . . . . . . . . . . . . . . . $\uparrow$ . . . . . . . . $\uparrow$
. . . . . . . . . . . . . . . . . . . . . . . . .
We have: . $(z_1)(z_2) \;=\;(\sqrt{6}\cdot\sqrt{2})\bigg[\cos\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) \bigg]$
. . . . . . . . . . . . . $= \;2\sqrt{3}\,\left(\cos\frac{25\pi}{12} + i\sin\frac{25\pi}{12}\right)$
. . . . . . . . . . . . . $=\;2\sqrt{3}\,\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$