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

Question

Please help me by showing alternate methods to solve this complex number SAT question

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?

Thank you in advance :)

Answer 1

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.

Answer 2

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.

Answer 3

I think the method you are using easier one.

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

Multiply and divide the fraction by conjugate of denominator,

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

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

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

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

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

Answer 4

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,

$$34b=-23.$$

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Exploiting the proposed answers:

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

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

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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.