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arg(z+2)=(pi/6) i cannot get the answer in my book of y=(x+2)/(sqrt(3)) can anyone show me there steps?

 Apr 24, 2014

Best Answer 

 #2
avatar+118703 
+5

I am just learning from Alan's post - Thank you Alan. 

arg(z+2)=π6arg(x+iy+2)=π6arg((x+2)+iy)=π6tan(π6)=yx+213=yx+2y=x+23

See the diagram at this address - I haven't work out how to insert pictures yet.

http://gyazo.com/7bb85148a5e4c5c1cb6bddbfc206bbea

 Apr 25, 2014
 #1
avatar+33654 
+5

Represent the complex number z by x + iy where x and y are real numbers. here, the arg function represents the angle between the line from the origin to the point {x+2, y} and the x-axis.  We are told this angle is pi/6. The tangent of this angle is just y/(x+2) so we know that:

tan(pi/6) = y/(x+2)

Now tan(pi/6) or tan(30°) is just 1/sqrt(3) so 1/sqrt(3) = y/(x+2)

Multiply both sides by x+2 to get y = (x+2)/sqrt(3)

 

tanofpiby6=tan360(30)=tanofpiby6=0.57735026919

oneonsqrt3=13=oneonsqrt3=0.5773502691896258

 Apr 24, 2014
 #2
avatar+118703 
+5
Best Answer

I am just learning from Alan's post - Thank you Alan. 

arg(z+2)=π6arg(x+iy+2)=π6arg((x+2)+iy)=π6tan(π6)=yx+213=yx+2y=x+23

See the diagram at this address - I haven't work out how to insert pictures yet.

http://gyazo.com/7bb85148a5e4c5c1cb6bddbfc206bbea

Melody Apr 25, 2014

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