dealing with -i

**GalenWilson** · May 14th 2008, 02:18 PM

I know its not acceptable to have i in the denominator of an equation, however I'm not sure what to do when I have -i, do I need to get rid of the - (negative sign) sign?

My example is that using the quadratic I've gotten to this point z= (–2 – i ± sqrt(3))/2i. I know I need to get rid of the i in the denominator so I multiply the whole thing by i/i then I get -i + 1/2 ± (i*sqrt(3))/2. I'm wondering if I need to get rid of the negative sign in front of the i or if I'm doing something wrong, or can i just leave it as is.

Thanks

**Mathstud28** · May 14th 2008, 02:23 PM

Originally Posted by GalenWilson

I know its not acceptable to have i in the denominator of an equation, however I'm not sure what to do when I have -i, do I need to get rid of the - (negative sign) sign?

My example is that using the quadratic I've gotten to this point z= (–2 – i ± sqrt(3))/2i. I know I need to get rid of the i in the denominator so I multiply the whole thing by i/i then I get -i + 1/2 ± (i*sqrt(3))/2. I'm wondering if I need to get rid of the negative sign in front of the i or if I'm doing something wrong, or can i just leave it as is.

Thanks

I cant really te,ll what equation that is but if it is helpful $-i=i\cdot{-1}=i\cdot{i^2}=i^3$

**GalenWilson** · May 14th 2008, 02:46 PM

The questions starts out as iz^2 + (2 + i)z + 1. Using the quadratic equation it becomes z= (-2 - i ± sqrt(3))/2i. I know I need to get rid of the i in the denominator by multiplying the whole problem by i/i. I *believe* this leaves me with -i + 1/2 ± (i*sqrt(3))/2.

My question is A) am I wrong and B) if not then what do I do (if anyhting with -i?

I think I'm just confused.

**topsquark** · May 14th 2008, 06:41 PM

Originally Posted by GalenWilson

The questions starts out as iz^2 + (2 + i)z + 1. Using the quadratic equation it becomes z= (-2 - i ± sqrt(3))/2i. I know I need to get rid of the i in the denominator by multiplying the whole problem by i/i. I *believe* this leaves me with -i + 1/2 ± (i*sqrt(3))/2.

My question is A) am I wrong and B) if not then what do I do (if anyhting with -i?

I think I'm just confused.

$\frac{-2 - i \pm \sqrt{3}}{2i}$

$= \frac{-2 - i \pm \sqrt{3}}{2i} \cdot \frac{i}{i}$

$= \frac{-2i - i^2 \pm i\sqrt{3}}{2i^2}$

$= \frac{-2i +1 \pm i\sqrt{3}}{-2}$

$= i - \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$

Do you need to get rid of the - when it's in the denominator? (ie. Must you do that last step?) No, but I'd say it's likely a preferred form.

-Dan

**Plato** · May 15th 2008, 01:06 PM

In general here is are very very useful facts about complex numbers:
$\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$ , so $\frac{1}{{a + bi}} = \frac{{a - bi}}{{a^2 + b^2 }}$ .

Examples: $\frac{1}{{ - 2 + 3i}} = \frac{{ - 2 - 3i}}{{13}}$
$\left( {4 - 5i} \right)^{ - 4} = \frac{1}{{\left( {4 - 5i} \right)^4 }} = \left[ {\frac{{4 + 5i}}{{41}}} \right]^4$

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