[SOLVED] Complex numbers?

**mark1950** · June 3rd 2009, 10:00 AM

Determine the value of a if (√3 + ai)/(1 - √3i) is a real number, and find this number. The answer they gave is a = 3 and the real number is √3.

If (√3 + ai)/(1 - √3i) is a real number, then doesn't it mean that (√3 + ai)/(1 - √3i) is x in x + yi? So, is it correct to rearrange it in this way,

(√3 + ai)/(1 - √3i) + yi = z

But then, if I arranged it in that way, I cannot see any possible way to find "a". Can you help me? Thanks.

**bobak** · June 3rd 2009, 10:11 AM

(delete)

**Amer** · June 3rd 2009, 10:18 AM

$\frac{\sqrt{3}+ai}{1-\sqrt{3}i}$

$\frac{\sqrt{3}+ai}{1-\sqrt{3}i}\times\frac{1+\sqrt{3}i}{1+\sqrt{3}i}$

now after you modify it you should but the value of a which make the number real (make the i(0))

**TheAbstractionist** · June 3rd 2009, 10:22 AM

Originally Posted by mark1950

Determine the value of a if (√3 + ai)/(1 - √3i) is a real number, and find this number. The answer they gave is a = 3 and the real number is √3.

If (√3 + ai)/(1 - √3i) is a real number, then doesn't it mean that (√3 + ai)/(1 - √3i) is x in x + yi? So, is it correct to rearrange it in this way,

(√3 + ai)/(1 - √3i) + yi = z

But then, if I arranged it in that way, I cannot see any possible way to find "a". Can you help me? Thanks.

Hi mark1950.

The answer they give should be $a=-3.$

What you want to do is to rationalize the denominator.

$\frac{\sqrt3+ai}{1-\sqrt3i}\ =\ \frac{(\sqrt3+ai)(1+\sqrt3i)}{(1-\sqrt3i)(1+\sqrt3i)}\ =\ \frac{\sqrt3-\sqrt3a+(a+3)i}{1+3}\ =\ \frac{\sqrt3-\sqrt3a}4+\frac{a+3}4i$

Now you set the imaginary part to be 0 and find $a,$ then substitute this into the real part to find the real number.

**Plato** · June 3rd 2009, 10:23 AM

$\frac{{\left( {\sqrt 3 + ai} \right)}} {{\left( {1 - \sqrt 3 i} \right)}} = \frac{{\left( {\sqrt 3 + ai} \right)\left( {1 + \sqrt 3 i} \right)}} {4} = \frac{{\left( {\sqrt 3 - a\sqrt 3 } \right) + i\left( {3 + a} \right)}} {4}$

From the above it is clear that $a=-3$ gives a real number.
So, the given answer in not correct.

**mark1950** · June 3rd 2009, 10:27 AM

Oh, thanks people! Many mighty thanks. Huh...should have asked you guys before I wasted half of my pen's ink on this question. Thanks again, dudes! By the way, what does it mean by (√3 + ai)/(1 - √3i) is a real number? If what you guys did above is true ( which I believe it really is) then, wouldn't it make (√3 + ai)/(1 - √3i) a complex number but not a real number? Because if it was a real number, then the equation (√3 + ai)/(1 - √3i) + yi = z would have been correct, right? Sorry, as you can see my tag name, I'm a newbie in polynomials.

And, the Abstractionist,

(a + 3)/4 = 0
a = -3

Why do I have to set the imaginary part to 0? Is there any condition that I missed out?

**Plato** · June 3rd 2009, 11:04 AM

Originally Posted by mark1950

Why do I have to set the imaginary part to 0? Is there any condition that I missed out?

Any complex number is made like this: $z = \Re (z) + i\Im (z)$ .
It consists of real and imaginary parts.
Here are some examples: $\begin{gathered} \Re (3 - 2i) = 3\;\& \,\Im (3 - 2i) = - 2 \hfill \\ \Re ( - 1 + 2i) = - 1\;\& \,\Im ( - 1 + 2i) = 2 \hfill \\ \end{gathered}$ .

Thus $\Im \left( {\frac{{\left( {\sqrt 3 - a\sqrt 3 } \right) + i\left( {3 + a} \right)}} {4}} \right) = \frac{{3 + a}}{4}$ .

A number is real if the imaginary part is zero.

**Amer** · June 3rd 2009, 11:04 AM

in general we write the complex number like

$z=x+iy$

when we say $\frac{\sqrt3+ai}{1-\sqrt3i}$

real number that's mean the factor of i should be zero because if we have a factor instead of zero that mean it is not a real number it is imaginary I think you know that

$i=(-1)^{\frac{1}{2}}$

for example if they ask you is $\frac{6+\sqrt{2}}{i-\sqrt{2}}$
real you should simplify make no i in the denominator bu multiply it with $i+\sqrt{2}$
like that we remove i from the denominator you will see that it is not a real because we find i in the number try to evaluate it

I think it is clear now

**mark1950** · June 3rd 2009, 06:15 PM

Thanks! Your description have cleared me doubts!

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