Jiya Sharma Posted April 23, 2014 Report Share Posted April 23, 2014 If Arg(z) = 0 , show that z is real and positive. Reply Link to post Share on other sites More sharing options...
racism Posted April 23, 2014 Report Share Posted April 23, 2014 jkl;jkl;yuio Reply Link to post Share on other sites More sharing options...
cricketcrazynerd Posted April 23, 2014 Report Share Posted April 23, 2014 This is best done using a diagram. If arg(z)=0, it means the angle between the complex point (the vector leading to it) and the x-axis is 0. This implies that the point lies just on the x-axis (that represents real numbers) and the positive side (the angle is 0, not 180). There is no imaginary part to the number. 3 Reply Link to post Share on other sites More sharing options...
Fiz Posted April 23, 2014 Report Share Posted April 23, 2014 Try to conceptualize it. An it is a concept number when drawn on an Argand's Diagram it will have an angle of 0.So its just a line that coincides the real part, so there is no imaginary part.This is just an idea of the solution.Good Luck! Reply Link to post Share on other sites More sharing options...
Negotiation Posted April 23, 2014 Report Share Posted April 23, 2014 If Arg(z) = 0 , show that z is real and positive. Let the modulus be a real number r. This must be real by the definition of modulus.Then, z = r cis(0) = r (cos(0) + i sin(0) ) = r (1 + 0i ) = r (1) = r.As r is real, z is real. 3 Reply Link to post Share on other sites More sharing options...
Thrashmaster Posted April 23, 2014 Report Share Posted April 23, 2014 If Arg(z) = 0 , show that z is real and positive. Let the modulus be a real number r. This must be real by the definition of modulus.Then, z = r cis(0) = r (cos(0) + i sin(0) ) = r (1 + 0i ) = r (1) = r.As r is real, z is real.We were just covering this in Math HL earlier this week! And if that's a real IB question the OP asked, that seems to be one of the easier complex number questions. Reply Link to post Share on other sites More sharing options...
Negotiation Posted April 25, 2014 Report Share Posted April 25, 2014 We were just covering this in Math HL earlier this week! And if that's a real IB question the OP asked, that seems to be one of the easier complex number questions.Are you still covering the syllabus? Reply Link to post Share on other sites More sharing options...
elaifyanre Posted April 25, 2014 Report Share Posted April 25, 2014 We were just covering this in Math HL earlier this week! And if that's a real IB question the OP asked, that seems to be one of the easier complex number questions.Are you still covering the syllabus?He is a may 2015 candidate. 2 Reply Link to post Share on other sites More sharing options...
Negotiation Posted April 25, 2014 Report Share Posted April 25, 2014 He is a may 2015 candidate.That makes it much better! 1 Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.