Complex number(exponential form)

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Homework Help Overview

The discussion revolves around finding a possible value for the complex variable w that satisfies the equation e^w = Z, where Z is defined as 1 + √3. Participants explore the exponential form of complex numbers and the relationships between their polar and rectangular representations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion of Z into exponential form and the implications of using polar coordinates. There are attempts to verify the correctness of the proposed solution and to clarify the relationship between the variables involved.

Discussion Status

Some participants have offered guidance on verifying the solution, and there appears to be a general agreement on the correctness of the approach taken. However, explicit consensus on the final answer has not been reached, as some participants continue to seek confirmation.

Contextual Notes

There is a mention of potential confusion regarding the representation of Z and the use of complex notation. Participants are also navigating the nuances of expressing complex numbers in different forms.

naspek
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Homework Statement



hey there.. I've try this question(below)..
but, i don't know whether my solution is correct or not..

Find a possible value for w that satisfies relation...
e^w = Z -->where z = 1 + 3^1/2


my solution...
r = 2
theta = pi/3
change z into exponential form...
z = 2e^i(pi/3)

hence, w = i(pi/3)

can anybody pleas guide me or verified my answer...
 
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You mean
R.e^w = Z -->where z = 1 + 3^1/2.j

If r = 2
and w = j (pi/3)

Knowing
R . e^(j.t) = R. [cos(t) + j sin(t)],

2 . e^(j. pi /3) = ... ?
 
rootX said:
2 . e^(j. pi /3) = ... ?

2 . e^(j. pi /3) = 2[cos (pi/3) + i sin (pi/3)]
 
naspek said:
2 . e^(j. pi /3) = 2[cos (pi/3) + i sin (pi/3)]

Which equals z, hence verifying that your solution is good.
 
ok.. so.. is my answer is correct?
 
rootX said:
...verifying that your solution is good.
naspek said:
ok.. so.. is my answer is correct?
What rootX said is another way of saying "yes, it is correct" :smile:
 
thanks guys! really appreciate your help.. =)
 

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