Which of the following complex numbers is equivalent to $\frac{3-5i}{8+2i}$?

I encountered this question in one of my SAT practice tests.

I know the answer is option C, however the only way I got the answer was by trial and error of trying multiple ways to simplify the equation and I ended up rationalising it to get answer choice C.

Is there any other, perhaps easier or more direct method, that I can use to solve these types of questions?

• Mutiplying top and bottom by the conjugate of the denominator is the standard approach. No need for trial and error -- the standard method will always work. – quasi Mar 29 '17 at 6:36
• I ended up rationalising it Don't know that there is anything more direct than that. You don't even need to carry out all calculations, just figure out the sign of the imaginary part to decide between C and D. (That's assuming you discarded choices A, B upfront, as expected) – dxiv Mar 29 '17 at 6:39
• As a multiple choice question, there will be a minus sign, but A is likely to be wrong – Henry Mar 29 '17 at 6:45
• As a multiple choice question ,you can compare $|z|=|\dfrac{3-5i}{8+2i}|=\dfrac{|3-5i|}{|8+2i|}=\dfrac{\sqrt{2(17)}}{\sqrt{4(17)}}=\dfrac{\sqrt2}{2}$ – Khosrotash Mar 29 '17 at 6:56
• @Khosrotash: interesting proposal, but unfortunately it leads to harder computation than the standard way (because of $(7^2+23^2)/34^2$) :-( – Yves Daoust Mar 29 '17 at 7:10

Two ways to approach this problem. First: As quasi suggests in the comments, multiply by the conjugate of the denominator.

\begin{align} \frac{3-5i}{8+2i} & = \frac{3-5i}{8+2i} \times \frac{8-2i}{8-2i} \\ & = \frac{24-40i-6i-10}{8^2+2^2} \\ & = \frac{14-46i}{68} = \frac{7-23i}{34} \end{align}

The second approach, given that it's a multiple choice problem, is to multiply each of the answers by $8+2i$ and see if you obtain $3-5i$. I think the first approach is simpler, but they'll both work.

Multiplying the top and bottom by the complex conjugate is how you handle this. That gives:

$$\frac{3-5i}{8+2i}\cdot\frac{8-2i}{8-2i}=\frac{(3-5i)(8-2i)}{8^2+2^2}=\frac{7-23i}{34}$$

Notice that at the second step we are guaranteed to have a real denominator because $(a+bi)(a-bi)=a^2+b^2$ is always real.

I think the method you are using easier one.

$$\frac{3-5i}{8+2i}$$

Multiply and divide the fraction by conjugate of denominator,

$$\frac{3-5i}{8+2i} \cdot \frac{8-2i}{8-2i}$$

$$\frac{24-6i-40i+10i^2}{64-4i^2}$$

$$\frac{24-6i-40i-10}{64+4}$$

$$\frac{14-46i}{68}$$

$$\frac{7-23i}{34}$$

Alternatives:

Let the answer be $a+ib$. Rewrite the initial equation as $$(a+ib)(8+i2)=3-i5.$$

Expanding, you get the $2\times2$ system $$\begin{cases} 8a-2b=3,\\ 2a+8b=-5.\end{cases}$$

Then by Cramer or simply adding four times the first equation and the second

$$34a=7$$

and subtracting the first from four times the second,

$$34b=-23.$$

Get rid of the denominators and try the products

$$(3\pm i20)(8+i2)=24\mp40+i(6\pm160)\to64-i154,$$ $$(7\pm i23)(8+i2)=56\mp46+i(14\pm144)\to102-i170.$$

(We select the signs that match those of $3-i5$.)

Then we obtain the identity

$$(7-i23)(8+i2)=34(3-i5).$$

The real way:

Use the division formula

$$\frac{a+ib}{c+id}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2},$$

giving

$$\frac{14}{68}-i\frac{46}{68}.$$

Then compare to the proposed answers. As C seems to match but D is similar, double check the signs or try the product, for safety.