Complex Number Question (1 Viewer)

amdspotter · Oct 20, 2021

If someone could assist me in how to do these two questions that would be greatly appreciated.
Thanks

jimmysmith560 · Oct 20, 2021

Would this help for 14?

amdspotter · Oct 20, 2021

jimmysmith560 said:
Would this help for 14?

View attachment 32847

Some sort of explanation for whats occuring/ what the solution did in line 1 would be helpful. Also thanks mate

notme123 · Oct 20, 2021

amdspotter said:
Some sort of explanation for whats occuring/ what the solution did in line 1 would be helpful. Also thanks mate

if you manipulate the numerator so it matches the denominator, you must also undo it to not change the question/identity its asking e.g. if you minus i you must also + i to cancel out and keep the q the same. he then simplifies and says if what he got is the same as the original thing that is given then since the numerator is real, for the fraction to be real, the denomiator must also be real, hence so is the fraction with z.

for 15 divide both sides by the lhs and convert z to rcis(t) and zbar to rcis(-t). in the end you should get cis(4t) = 1 to which the solutions are cos(4t)=1 or in other words, 4t=0,2pi,-2pi,4pi,-4pi
therefore the only values t can have are t = k(pi/2) where k is an integer including 0. this means z can only be real or only imaginary

yanujw · Oct 20, 2021

15.
$Let z = a +ib$
$\overline{z} = a-ib$
$a^2 + 2abi - b^2 = a^2 - 2abi - b^2$
$4abi = 0$
By null factor law, a = 0 or b = 0

If a = 0, $z = bi$ and is purely imaginary
If b = 0, $z = a$ and is purely real

Velocifire · Oct 20, 2021

yanujw said:
15.
$Let z = a +ib$
$\overline{z} = a-ib$
$a^2 + 2abi - b^2 = a^2 - 2abi - b^2$
$4abi = 0$
By null factor law, a = 0 or b = 0

If a = 0, $z = bi$ and is purely imaginary
If b = 0, $z = a$ and is purely real

unrelated and don't do 4u myself, but that signature slaps hard

Gotta get out of here

CM_Tutor · Oct 20, 2021

I think, with q14, what they are expecting is for you to put $z=a+bi$ into $\cfrac{z}{z-i}$ , as in

$\begin{align*} \cfrac{z}{z-i} &= \cfrac{a + ib}{a+ib-i} \\ &= \cfrac{a + ib}{a+i(b-1)} \times \cfrac{a - i(b-1)}{a - i(b-1)} \\ &= \cfrac{a^2 - ai(b - 1) + iba - i^2b(b - 1)}{a^2 + i^2(b-1)^2} \\ &= \cfrac{a^2 + b(b - 1) + ai}{a^2 - (b-1)^2} \\ \text{Now, since $\cfrac{z}{z-i}$ is real:} \qquad \Im{\left(\cfrac{z}{z-1}\right)} &= 0 \\ \cfrac{a}{a^2 - (b-1)^2} &= 0 \\ a &= 0 \qquad \text{in which case $z$ is purely imaginary} \\ \text{And,} \qquad \cfrac{z}{z-i} &= \cfrac{0^2 +b(b-1) + 0}{0^2-(b-1)^2} \\ &= \cfrac{b}{1-b} \qquad \text{which is real for all $b$ except $b = 1$} \end{align*}$

So, $\cfrac{z}{z-i}$ is undefined if $z = i$ and is real so long as $z$ is 0 or purely imaginary.

CM_Tutor · Oct 20, 2021

For q 15, you can assume that $z \ne 0$ (the result is trivially true if $z=0$ ) and then use:

$\begin{align*} z\bar{z} &=|z|^2 \\ z^2{\bar{z}}^2 &= |z|^4 \\ \text{Since we know that $z^2 = {\bar{z}}^2$:} \qquad z^2 \times z^2 &= |z|^4 \\ z^4 &= k^4 \qquad \text{where $k = |z|$, a positive real} \\ z^2 &= \pm k^2 \\ \text{So,} \qquad z &= \pm ki\ \text{or}\ \pm k \qquad \text{making $z$ purely real or purely imaginary} \end{align*}$

amdspotter · Oct 20, 2021

Just opened bos and saw many answers. Thanks a lot @CM_Tutor @yanujw really appreciate it. And yep through how U guys have explained it I understand it now, thx once again

CM_Tutor · Oct 20, 2021

Another approach for q15:

$\begin{align*} z^2 &= {\bar{z}}^2 \\ z &= \pm\bar{z} \\ \\ \text{If}\ z &= \bar{z}\ \text{then}\ \Im{(z)} = 0 \qquad \implies \qquad \text{$z$ is purely real} \\ \\ \text{If}\ z &= -\bar{z} \\ z + \bar{z} &= 0\ \text{then}\ \Re{(z)} = 0 \qquad \implies \qquad \text{$z$ is purely imaginary} \end{align*}$

Drdusk · Oct 20, 2021

amdspotter said:
View attachment 32846
If someone could assist me in how to do these two questions that would be greatly appreciated.
Thanks

Alternatively you could say for 14) that because z/(z-i) is real,
$\arg\bigg(\dfrac{z}{z-i}\bigg) = 0$

$\therefore \arg(z) - \arg(z-i) = 0 \Rightarrow \fbox{arg(z) = arg(z-i)}$

You could then argue that the above equality is only true if z does NOT have a real part, or of course if it's 0.

amdspotter · Oct 20, 2021

Drdusk said:
Alternatively you could say for 14) that because z/(z-i) is real,
$\arg\bigg(\dfrac{z}{z-i}\bigg) = 0$

$\therefore \arg(z) - \arg(z-i) = 0 \Rightarrow \fbox{arg(z) = arg(z-i)}$

You could then argue that the above equality is only true if z does NOT have a real part, or of course if it's 0.

Ah ok this makes sense, thanks

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Complex Number Question (1 Viewer)

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