Imaginary Numbers

Given that Z=(a+i)^4 where a is real, find values for a such that
(i) Z is real
(ii) Z is wholly imaginary

Not sure how to start either of these off, any tips?

Thanks

Start by expanding (a+i)^4. Have you tried this?

I got a^4 + 4ia^3 -6a^2 -4ia +1 but don’t see what to do from here.

Next you need to write the expanded $z$ so that it's in the form $x iy$, in other words you need to separate the real and imaginary parts. This is a common technique in this topic.

So by separating the $i$ terms and factorising you have $z = (a^4 - 6a^2 1) (4a^3-4a)i$

Now the real part and imaginary part are clearly shown. If $z$ is real, what can you say about the complex number above?

As Notnek stated earlier, upon expanding, you're looking to force a condition on it in the two parts. Writing it in form x+iy, to seperate real and imaginary parts.

Part i) Z being real, meaning a complex part of 0, given your expression above, and the condition given, how can you go about finding out.

Part ii) Z being wholly imaginary, means a real part of 0, again, another condition, given your real and imaginary parts, similarly to part i), can you see how to go around solving it?

Sorry but I really don’t know what to say. Is it to do with the last bracket? Does 4a^3 -4a=0 since i*0 would disregard that bracket so Z would be real? a=0,1,-1?

Yes that’s correct. Now try the next part.

Would it be the same but this time the other bracket will equal zero so we are just left with the imaginary part? a^4 -6a^2 +1= 0 therefore using a^2=x we can solve like a quadratic. Using the formula I get a= 0.41... or 2.41....

It doesnt factorise so you can just complete the square to get $(a^2-3)^2-8=0$

Hence $a^2 = 3 \pm 2\sqrt{2}$. Now notice that $3 \pm 2\sqrt{2} > 0$, so there will be 4 possibilites for $a$.

Not entirely sure what you mean but I got a=0.41... and a=2.41... by square rooting as normal but not sure how you get the other answers? Are they the same but -0.41... and -2.41... instead?

Yes.

If $a^2 = 3+2\sqrt{2}$ then clearly $a = \pm \sqrt{3+2\sqrt{2}} \approx \pm 2.4142$.

Likewise if $a^2 = 3-2\sqrt{2}$.