Answer in polar form eter to quoTy eiolim o eilbZ,= 13+i z,-51Z-512iZ, =512-5/2iol-oao-bas-8-ow1 20) mrequoltZiZ2 =%3DeoldshaV lexov2 ni raMos stonob owt bn soln sno.otos owt obno snnA goile softool nbo vds r2 aa bna hmob oondi br oloo sno25.012 agmoldonq i ovlor ot ysoon anoiisups11odmapote22asldahsys0 basasoiuj owi pofteo ono obno oda li yoq ol ovd522E--AAxhtsm lo nsmmatob

The given equations are: z1=3+i z2=52-52i General representation of complex number…

Answered: limto Z,=V3+i z,-51Z-512i ZiZ2 =…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Answer in polar form

eter to quoTy ei
olim o eilb
Z,= 13+i z,-51Z-512i
Z, =512-5/2i
ol-oao-bas-
8-ow1 20) mr
equ
olt
ZiZ2 =
%3D
eoldshaV lexov2 ni ra
Mos s
tonob owt bn soln sno.otos owt obno snnA goile softo
ol nbo vds r2 aa bna hmob oondi br oloo sno
25.012 ag
moldonq i ovlor ot ysoon anoiisups
11
odmapote
22
asldahsy
s0 basasoiuj owi pofteo ono obno oda li yoq ol ovd
522
E--AAxhtsm lo nsmmatob

Expert Solution

Step 1

The given equations are:

$z_{1} = \sqrt{3} + i$

$z_{2} = 5 \sqrt{2} - 5 \sqrt{2} i$

General representation of complex number is:

$a + i b$

General representation of complex number into polar form is:

$a + i b$ = $r (\cos θ + i \sin θ)$ ….(1)

Here, $r = \sqrt{a^{2} + b^{2}}$ and $θ = \tan^{- 1} \frac{b}{a}$

Polar form of $z_{1} z_{2}$ is given as:

$\begin{array}{rcl} z_{1} z_{2} & = & (\sqrt{3} + i) (5 \sqrt{2} - 5 \sqrt{2} i) \\ = & 5 \sqrt{6} - 5 \sqrt{6} i + 5 \sqrt{2} i + 5 \sqrt{2} \\ = & (5 \sqrt{6} + 5 \sqrt{2}) + (5 \sqrt{2} - 5 \sqrt{6}) i \end{array}$

$\begin{array}{rcl} r & = & \sqrt{({(5 \sqrt{6} + 5 \sqrt{2})}^{2} + {(5 \sqrt{2} - 5 \sqrt{6})}^{2})} \\ = & \sqrt{(200 + 50 \sqrt{2} + 200 - 50 \sqrt{2})} \\ = & \sqrt{400} \\ = & 20 \end{array}$

$θ = \tan^{- 1} \frac{5 \sqrt{2} - 5 \sqrt{2} i}{\sqrt{3} + i} = \frac{- π}{12}$

Substitute the value of $r$ and $θ$ in equation (1)

$(5 \sqrt{6} + 5 \sqrt{2}) + (5 \sqrt{2} - 5 \sqrt{6}) i = 20 (\cos \frac{- π}{12} + i \sin \frac{- π}{12})$

Therefore, the polar form of $z_{1} z_{2} = 20 (\cos \frac{- π}{12} + i \sin \frac{- π}{12})$

Polar form of $\frac{z_{1}}{z_{2}}$ is given as:

$\begin{array}{rcl} \frac{z_{1}}{z_{2}} & = & \frac{\sqrt{3} + i}{5 \sqrt{2} - 5 \sqrt{2} i} \\ = & \frac{(\sqrt{3} + i) (5 \sqrt{2} + 5 \sqrt{2} i)}{(5 \sqrt{2} - 5 \sqrt{2} i) (5 \sqrt{2} + 5 \sqrt{2} i))} \\ = & \frac{(5 \sqrt{6} - 5 \sqrt{2}) + (5 \sqrt{2} + 5 \sqrt{6}) i}{{(5 \sqrt{2})}^{2} + {(5 \sqrt{2})}^{2}} \\ = & \frac{(5 \sqrt{6} - 5 \sqrt{2}) + (5 \sqrt{2} + 5 \sqrt{6}) i}{100} \end{array}$

$= \frac{\sqrt{6} - \sqrt{2}}{20} + i \frac{(\sqrt{2} + \sqrt{6})}{20}$

$\begin{array}{rcl} r & = & \sqrt{{(\frac{\sqrt{6} - \sqrt{2}}{20})}^{2} + {(\frac{\sqrt{2} + \sqrt{6}}{20})}^{2}} \\ = & \frac{1}{5} \end{array}$

$\begin{array}{rcl} θ & = & \tan^{- 1} \frac{\frac{\sqrt{2} + \sqrt{6}}{20}}{\frac{\sqrt{6} - \sqrt{2}}{20}} \\ = & \frac{5 π}{12} \end{array}$

Substitute the above values of $r$ and $θ$ in equation (1)

$\frac{\sqrt{6} - \sqrt{2}}{20} + i \frac{\sqrt{2} + \sqrt{6}}{20} = \frac{1}{5} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$

Therefore, the polar form of $\frac{z_{1}}{z_{2}} = \frac{1}{5} (\cos \frac{5 π}{12} + i \sin \frac{5 π}{12})$ .

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