Complex numbers- locus (1 Viewer)

[HeroicPandas]

Let l be the line in the complex plane that passes through the origin and makes an angle alpha with the positive axis, where 0 < alpha < PI/2

 

[iampeterr]

what do you want to do with it ?

 

[HeroicPandas]

what do you want to do with it ?

oops sorry i forgot to write the next part of the question

...

The point P represents the complex number Z1, where 0 < arg (Z1) < alpha. the point P is reflected in the line l to produce the point Q, which represents the complex number Z2. hence |Z1| = |Z2|


a. explain why arg(Z1) + arg (Z2) = 2alpha

b. deduce that Z1 x Z2 = |Z1|^2 [cos(2alpha) + isin(2alpha)]

c. let alpha = PI/4 and let R be the point that represents the number Z1.Z2

describe the locus of R as Z1 varies

 

[iampeterr]

a) see that diagram, ill show u the common sense way.

Let arg z1 = theta therefore angle POL = alpha - theta

then construct a triangle POQ since |z1|=|z2| its an isoscles.

therefore arg z2 = alpha- theta +alpah = 2alpha - theta

therefore arg z1 + arg z2 = theta + 2alpha - theta = 2alpah =]

there are alot of methods, this is the way i see it.

b) we know z1 = |z1|(cos theta + isin theta) = |z1|cis(theta)
and z2= |z2|(cos 2alpha-theta +i sin 2alpha-theta ) but we know that |z1|=|z2|

there fore z1.z2 = |z1|cis(theta) * |z1| cis (2alpha - theta)

there fore we add arguments because its multiply...

there fore |z1|^2 cis (2alpha) = z1*z2=|z1|^2(cos2a+isin2a)

c) just sub in pi/4 into z1*z2 ..

 

[umm what]

oops sorry i forgot to write the next part of the question

...

The point P represents the complex number Z1, where 0 < arg (Z1) < alpha. the point P is reflected in the line l to produce the point Q, which represents the complex number Z2. hence |Z1| = |Z2|


a. explain why arg(Z1) + arg (Z2) = 2alpha

b. deduce that Z1 x Z2 = |Z1|^2 [cos(2alpha) + isin(2alpha)]

c. let alpha = PI/4 and let R be the point that represents the number Z1.Z2

describe the locus of R as Z1 varies

LMAO!

 

[HeroicPandas]

a) see that diagram, ill show u the common sense way.

Let arg z1 = theta therefore angle POL = alpha - theta

then construct a triangle POQ since |z1|=|z2| its an isoscles.

therefore arg z2 = alpha- theta +alpah = 2alpha - theta

therefore arg z1 + arg z2 = theta + 2alpha - theta = 2alpah =]

there are alot of methods, this is the way i see it.

b) we know z1 = |z1|(cos theta + isin theta) = |z1|cis(theta)
and z2= |z2|(cos 2alpha-theta +i sin 2alpha-theta ) but we know that |z1|=|z2|

there fore z1.z2 = |z1|cis(theta) * |z1| cis (2alpha - theta)

there fore we add arguments because its multiply...

there fore |z1|^2 cis (2alpha) = z1*z2=|z1|^2(cos2a+isin2a)

c) just sub in pi/4 into z1*z2 ..

BEAST, thank you a lot