If z1=r1(cosθ1+i sinθ1) and z2=r2(cosθ2+i sinθ2) are complex numbers, then (z1)/(z2), z2 ≠0, equ

Name: Problem 12, Chapter 10: Algebra and Trigonometry
Uploaded: 2020-06-17T14:29:27-08:00
Duration: 1 min 18 s
Channel: Joseph Lentino
Description: If z1=r1(cosθ1+i sinθ1) and z2=r2(cosθ2+i sinθ2) are complex numbers, then (z1)/(z2), z2 ≠0, equals which of the following? (A) (r1)/(r2)[cos(θ1-θ2)+i sin(θ1-θ2)] (B) (r1)/(r2)[cos((θ1)/(θ2))+i sin((θ1)/(θ2))] (C) (r1)/(r2)[cos(θ1+θ2)-i sin(θ1+θ2)] (D) (r1)/(r2)[cos((θ1)/(θ2))-i sin((θ1)/(θ2))]

step 1 So we're given our first polar point as r1, cosine a theta 1, plus i. sine a theta 1, which is t

Question

If $z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)$ and $z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)$ are complex numbers, then $\frac{z_{1}}{z_{2}}, z_{2} \neq 0,$ equals which of the following? (A) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right]$ (B) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)+i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$ (C) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}+\theta_{2}\right)-i \sin \left(\theta_{1}+\theta_{2}\right)\right]$ (D) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)-i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$

   If $z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)$ and $z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)$
are complex numbers, then $\frac{z_{1}}{z_{2}}, z_{2} \neq 0,$ equals which of the following?
(A) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right]$
(B) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)+i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$
(C) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}+\theta_{2}\right)-i \sin \left(\theta_{1}+\theta_{2}\right)\right]$
(D) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)-i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$

Algebra and Trigonometry

Michael Sullivan 10th Edition

Chapter 10, Problem 12

Instant Answer

Solved by Expert Joseph Lentino

06/17/2020

Step 1

The polar form of a complex number is given by $z=r(\cos \theta+i \sin \theta)$ where $r$ is the magnitude and $\theta$ is the angle of rotation. Show more…

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If $z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right)$ and $z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right)$ are complex numbers, then $\frac{z_{1}}{z_{2}}, z_{2} \neq 0,$ equals which of the following? (A) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}-\theta_{2}\right)+i \sin \left(\theta_{1}-\theta_{2}\right)\right]$ (B) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)+i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$ (C) $\frac{r_{1}}{r_{2}}\left[\cos \left(\theta_{1}+\theta_{2}\right)-i \sin \left(\theta_{1}+\theta_{2}\right)\right]$ (D) $\frac{r_{1}}{r_{2}}\left[\cos \left(\frac{\theta_{1}}{\theta_{2}}\right)-i \sin \left(\frac{\theta_{1}}{\theta_{2}}\right)\right]$

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Key Concepts

Polar Form Representation

Complex numbers can be expressed in polar form as r(cos ? + i sin ?), where r is the magnitude (or modulus) of the complex number, representing the distance from the origin in the complex plane, and ? is the argument, which is the angle formed with the positive real axis. This representation is beneficial as it simplifies many operations, such as multiplication, division, and finding powers and roots of complex numbers.

Division in Polar Form

When dividing two complex numbers expressed in polar form, the operation involves dividing their magnitudes and subtracting their angles. Specifically, if z1 = r1(cos ?1 + i sin ?1) and z2 = r2(cos ?2 + i sin ?2), then their quotient is given by (r1/r2)[cos(?1 ? ?2) + i sin(?1 ? ?2)]. This property is a direct result of the multiplicative properties of the modulus and the additive properties of the arguments in the polar representation.

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