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I am a bit confused about this question:

I understand that a purely real complex number is where the 'a' value is 0,

But how would you go about solving this question? So after you multiply it by the conjugate, would you make it = 0? If so, why?

Thanks a lot

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5 Answers

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Hint: Write $$\dfrac{k+4i}{1+ki}\times\dfrac{1-ki}{1-ki}$$

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In general complex number are in the form

$$z=x+iy$$

when $\Im(z)=y=0$ we have $z=x=\Re(z)$ and we talk of purely real number.

In other words in the complex plane $z$ lies on the $x$ axis.

Similarly when $x=0$ we have $z=y=\Im(z)$ and we talk of purely imaginary number.

In that case the condition is equivalent to

$$u=\bar u \iff \frac{k+4i}{1+ki}=\frac{k-4i}{1-ki}$$

that is

$$(k+4i)(1-ki)=(k-4i)(1+ki)$$

$$k-k^2i+4i+4k=k+k^2i-4i+4k$$

$$2k^2i=8i \iff k^2=4 \iff k=\pm 2$$

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$$u = \frac{k+4i}{1+ki}$$ We multiply the top and bottom of the fraction by the complex conjugate of the denominator, $(1-ki)$. This gives: $$u = \frac{k+4i}{1+ki} \frac{1-ki}{1-ki} = \frac{k-k^2i +4i +4k}{1+k^2}$$ We want all the terms with an $i$ to cancel. So, that is the same as finding the $k$ value that makes $-k^2i +4i = 0$. These values are only $k = \pm 2$.

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Assuming that $k$ is a real number ,$\frac {k+4i} { 1+ki}=\frac {(k+4i)(1-ki)} {(1+ki)(1-ki)}=\frac {4+4k+4i-k^{2}i} {1+k^{2}}$ so the imaginary part is $0$ iff $k^{2}=4$ or $k =\pm 2$. ,

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A complex number $u=a+bi$ will be a purely real number if $b=0$. Then: $$u = \frac{k+4i}{1+ki}=a \iff k+4i=a+aki \iff (a-k)+(4-ak)i=0 \iff \\ \begin{cases}a-k=0 \\ 4-ak=0\end{cases} \iff 4-k^2=0 \iff k=\pm 2. $$

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