Q. 66 Prove that z1z2=r1r2(cos(胃1-胃2... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 9: Q. 66 (page 592)

Prove that $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} [(\cos (胃_{1} - 胃_{2}) + i \sin (胃_{1} - 胃_{2})]$

Short Answer

Expert verified

Assume two complex numbers and then divide both of them. Then rationalise the denominator we can get the formula

$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} [(\cos (胃_{1} - 胃_{2}) + i \sin (胃_{1} - 胃_{2})]$

Step by step solution

Step 1. Considering two complex numbers

First we consider two complex numbers

$z_{1} = r_{1} [\cos 胃_{1} + i \sin 胃_{1}] z_{2} = r_{2} [\cos 胃_{2} + i \sin 胃_{2}]$

Step 2. Calculations

Now we divide $z_{1}$ by $z_{2}$

role="math" localid="1646739281844" $\frac{z_{1}}{z_{2}} = \frac{r_{1} [\cos 胃_{1} + i \sin 胃_{1}]}{r_{2} [c o s 胃_{2} + i \sin 胃_{2}]}$

Now multiplying the conjugate of the denominator to the numerator and the denominator.

$\frac{z_{1}}{z_{2}} = \frac{r_{1} [\cos 胃_{1} + i \sin 胃_{1}]}{r_{2} [c o s 胃_{2} + i \sin 胃_{2}]} (\frac{[\cos 胃_{2} - i \sin 胃_{2}]}{[c o s 胃_{2} - i \sin 胃_{2}]}) = \frac{r_{1}}{r_{2}} \frac{(\cos 胃_{1} + i \sin 胃_{1}) (\cos 胃_{2} - i \sin 胃_{2})}{\cos^{2} 胃_{2} + \sin^{2} 胃_{2}} = \frac{r_{1}}{r_{2}} (\cos 胃_{1} + i \sin 胃_{1}) (\cos 胃_{2} - i \sin 胃_{2}) = \frac{r_{1}}{r_{2}} (\cos 胃_{1} \cos 胃_{2} - \cos 胃_{1} i \sin 胃_{2} + i \sin 胃_{1} \cos 胃_{2} + \sin 胃_{1} \sin 胃_{2}) = \frac{r_{1}}{r_{2}} [(\cos 胃_{1} \cos 胃_{2} + \sin 胃_{1} \sin 胃_{2}) + i (\sin 胃_{1} \cos 胃_{2} - \cos 胃_{1} i \sin 胃_{2})] = \frac{r_{1}}{r_{2}} [\cos (胃_{1} - 胃_{2}) + i \sin (胃_{1} - 胃_{2})]$

Hence proved

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91影视

Prove that $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} [(\cos (胃_{1} - 胃_{2}) + i \sin (胃_{1} - 胃_{2})]$

Short Answer

Step by step solution

Step 1. Considering two complex numbers

Step 2. Calculations

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91影视

Prove that z1z2=r1r2(cos(胃1-胃2)+isin(胃1-胃2)

Short Answer

Step by step solution

Step 1. Considering two complex numbers

Step 2. Calculations

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Discrete Mathematics

Decision Maths

Statistics

Applied Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Prove that $\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} [(\cos (胃_{1} - 胃_{2}) + i \sin (胃_{1} - 胃_{2})]$