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Square root of complex number

  • #1

Homework Statement


show that [tex]\sqrt{1+ja}[/tex] is equivalent to [tex]\pm(1+j)(a/2)^{1/2}[/tex] with a>>1


Homework Equations


Euler's formula?


The Attempt at a Solution


with a>>1
|z| = [tex]\sqrt{(1 + a^{2})}[/tex] == a
lim a-->infinity arctan (a/1) == [tex]\pi/2[/tex]
[tex]\sqrt{z}[/tex] = [tex]\sqrt{(ae^{j\pi/2})}[/tex]
[tex]\sqrt{z}[/tex] = [tex]\pmsqrt a^{1/2} e^{j1.25}[/tex]

However, when I transfer back to complex form, I don't get it to equal 1+j. Not too sure how they got a/2 as well.

Any tips would be great.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
Where did you get 1.25 from, in the last line? That looks wrong...

I actually wouldn't bother with converting to decimals at all, since it would be inexact. Just leave it as [itex]e^{j\pi/2}[/itex]. What is the square root of that number?
 
  • #3
That came from the square root of pi / 2.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955

Homework Statement


show that [tex]\sqrt{1+ja}[/tex] is equivalent to [tex]\pm(1+j)(a/2)^{1/2}[/tex] with a>>1
You can't, it's not true. [itex][(1+ j)(a/2)^{1/2}]^2= (1+j)^2 (a/2)= (1+ 2j- 1)(a/2)= aj[/itex], not 1+ja. Of course, if, by "a>>1" you mean a is extremely large, then a is large compared to 1 so that is approximately true. Is that what you meant by "equivalent"?


Homework Equations


Euler's formula?


The Attempt at a Solution


with a>>1
|z| = [tex]\sqrt{(1 + a^{2})}[/tex] == a
lim a-->infinity arctan (a/1) == [tex]\pi/2[/tex]
[tex]\sqrt{z}[/tex] = [tex]\sqrt{(ae^{j\pi/2})}[/tex]
[tex]\sqrt{z}[/tex] = [tex]\pmsqrt a^{1/2} e^{j1.25}[/tex]

However, when I transfer back to complex form, I don't get it to equal 1+j. Not too sure how they got a/2 as well.

Any tips would be great.

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #5
diazona
Homework Helper
2,175
6
That came from the square root of pi / 2.
There was no reason to take the square root of [itex]\pi/2[/itex]. Do you remember how square roots act on exponentials? That is, in general, what's the square root of [itex]e^x[/itex]?

HallsofIvy makes a good point, too - I didn't realize that at first.
 
  • #6
You can't, it's not true. [itex][(1+ j)(a/2)^{1/2}]^2= (1+j)^2 (a/2)= (1+ 2j- 1)(a/2)= aj[/itex], not 1+ja. Of course, if, by "a>>1" you mean a is extremely large, then a is large compared to 1 so that is approximately true. Is that what you meant by "equivalent"?
Yeah I probably should have said approximately instead of equivalent. The question I posted is word for word out of an applied electromagnetism book for engineers. When it says assume a>>1, I just assumed a to be infinity.
 
  • #7
There was no reason to take the square root of [itex]\pi/2[/itex]. Do you remember how square roots act on exponentials? That is, in general, what's the square root of [itex]e^x[/itex]?

HallsofIvy makes a good point, too - I didn't realize that at first.
WHOOPS, I meant to type 0.785 instead of 1.25. I guess thats what I get for trying to simplify without looking at my hand written work. In either case, it still doesn't seem to work out. I'm kind of stumped as how to prove the (a/2)^1/2 part.
 

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