Proving the Equivalence of Square Root of Complex Numbers

[elimenohpee]

Homework Statement

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

Homework Equations

Euler's formula?

The Attempt at a Solution

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

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.

 

[diazona]

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 $e^{j\pi/2}$. What is the square root of that number?

 

[elimenohpee]

That came from the square root of pi / 2.

 

[HallsofIvy]

Homework Statement

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

You can't, it's not true. $[(1+ j)(a/2)^{1/2}]^2= (1+j)^2 (a/2)= (1+ 2j- 1)(a/2)= aj$, 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| = $$\sqrt{(1 + a^{2})}$$ == a
lim a-->infinity arctan (a/1) == $$\pi/2$$
$$\sqrt{z}$$ = $$\sqrt{(ae^{j\pi/2})}$$
$$\sqrt{z}$$ = $$\pmsqrt a^{1/2} e^{j1.25}$$

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.

 

[diazona]

That came from the square root of pi / 2.

There was no reason to take the square root of $\pi/2$. Do you remember how square roots act on exponentials? That is, in general, what's the square root of $e^x$?

HallsofIvy makes a good point, too - I didn't realize that at first.

 

[elimenohpee]

You can't, it's not true. $[(1+ j)(a/2)^{1/2}]^2= (1+j)^2 (a/2)= (1+ 2j- 1)(a/2)= aj$, 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.

 

[elimenohpee]

There was no reason to take the square root of $\pi/2$. Do you remember how square roots act on exponentials? That is, in general, what's the square root of $e^x$?

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 that's 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.