Given z1=23(cos25°+isin25°)z_1=\frac23\left(\cos25\degree+i\sin25\degree\right) and z2=52(cos15°+isin15°)z_2=\frac52\left(\cos15\degree+i\sin15\degree\right)z2=25(cos15°+isin15°), find the product z1・z2z_1・z_2z1・z2.

z1⋅z2=53CiS(40°)$$z_1$$\cdot $$z_2$$=\frac53CiS\left(40\degree\right)z1⋅z2=35CiS(40°)

Given z 1 = 2 3 ( cos 25 ° + i sin 25 ° ) z_1=\frac23\left(\cos25\degree+i\sin25\degree\right) and z 2 = 5 2 ( cos 15 ° + i sin 15 ° ) z_2=\frac52\left(\cos15\degree+i\sin15\degree\right) z 2 = 2 5 ( cos 15° + i sin 15° ) , find the product z 1 ・ z 2 z_1・z_2 z 1 ・ z 2 .

In this problem, we are given 2 complex numbers, z 1 and z 2. Both of these are in polar form, and we're asked to find their product. Now to find the product of z 1 and z 2, there are a couple steps that I need to take. The first step is going to be to take the r values out in front and multiply them. So you can see the r values are 2 thirds and 5 halves. So I'm going to have 2 thirds multiplied by 5 halves, that's multiplying the r values. Then what I need to do is I need to deal with these cosines and sines. Now recall that the cosine isine can be abbreviated to cis. So So we're going to have this multiplied by cis, and that's going to be the sum of the two angles. I can see that the first angle is 25 degrees. I can see that the second angle is 15 degrees. So this is what we end up getting. Now to solve the problem from here, well, all I really need to do is simplify everything that I have. I can simplify these 2 fractions by recognizing this 2 on top will cancel with that 2 on bottom. Meaning, all I'm gonna have is 5 thirds out in front. That's gonna be 5 thirds cis, and then we have 25 +15, which is 40 degrees. So this right here is the solution to z one times z 2, and that is how you can find the product of these two complex numbers. So hope you found this video helpful. Thanks for watching.

Given z1=23(cos⁡25°+isin⁡25°)$$z_1$$=\frac23\left(\cos25\degree+i\sin25\degree\right) and z2=52(cos⁡15°+isin⁡15°)$$z_2$$=\frac52\left(\cos15\degree+i\sin15\degree\right)z2=25(cos15°+isin15°), find the product z1・z2z_1・z_2z1・z2.

z1⋅z2=53CiS(40°)$$z_1$$\cdot $$z_2$$=\frac53CiS\left(40\degree\right)z1⋅z2=35CiS(40°)

z1・z2=53CiS(375°)$$z_1$$・$$z_2$$=\frac53CiS\left(375\degree\right)z1・z2=35CiS(375°)

z1・z2=196CiS(40°)$$z_1$$・$$z_2$$=\frac{19}{6}CiS\left(40\degree\right)z1・z2=619CiS(40°)

z1・z2=196CiS(375°)$$z_1$$・$$z_2$$=\frac{19}{6}CiS\left(375\degree\right)z1・z2=619CiS(375°)

8. Polar Coordinates and Complex Numbers

Products and Quotients of Complex Numbers

Algebra and Trigonometry

8. Polar Coordinates and Complex Numbers

Products and Quotients of Complex Numbers

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Multiple Choice

Given $z_{1} = \frac{2}{3} (\cos 25 ° + i \sin 25 °)$ and $z_2=$\frac$52$\left$($\cos$15$\degree$+i$\sin$15$\degree\]\right$)$ , find the product $z_1・z_2$ .

$z_1・z_2=$\frac$53CiS$\left$(375$\degree\]\right$)$

$z_1$\cdot$ z_2=$\frac$53CiS$\left$(40$\degree\]\right$)$

$z_1・z_2=$\frac{19}{6}$CiS$\left$(40$\degree\]\right$)$

$z_1・z_2=$\frac{19}{6}$CiS$\left$(375$\degree\]\right$)$

3:40m

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Struggling with Algebra and Trigonometry?

Given z1=23(cos25°+isin25°)z_1=\(\frac\)23\(\left\)(\(\cos\)25\(\degree\)+i\(\sin\)25\(\degree\]\right\)) and z2=52(cos15°+isin15°)z_2=\(\frac\)52\(\left\)(\(\cos\)15\(\degree\)+i\(\sin\)15\(\degree\]\right\))z2=25(cos15°+isin15°), find the product z1・z2z_1・z_2z1・z2.

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Given $z_{1} = \frac{2}{3} (\cos 25 ° + i \sin 25 °)$ and $z_2=\(\frac\)52\(\left\)(\(\cos\)15\(\degree\)+i\(\sin\)15\(\degree\]\right\))$ , find the product $z_1・z_2$ .