If$$z_1=2+3i$$$$z_2=1+2i$$then
$$z_1z_2-z_1^3=?$$

Question

If$$z_1=2+3i$$$$z_2=1+2i$$then
$$z_1z_2-z_1^3=?$$

empty · Answer

Give it your best try and show your steps

jiteshmeghwal9 · Answer

$$z_1.z_2=(2+3i)(1+2i)$$$$z_1z_2=(2-6)+i(4+3)$$$$z_1z_2=-4+7i$$now$$z_1^3=8+27i^3+54i^2+36i$$$$z_1^3=-46+27i^3+36i$$$$z_1z_2-z_1^3=42-29i+27i^3$$

jiteshmeghwal9 · Answer

now wht next ?

ganeshie8 · Answer

Nice! $i^3$ simplifies further right ?

empty · Answer

It looks almost right, except you seem to have messed up your terms here:

$z_1^3 = 8 + 12i + 18i^2 + 27i^3$

Which can also be simplified further with @ganeshie8 's comment. :D

jiteshmeghwal9 · Answer

$42-29i-27i$=$42-56i$

ganeshie8 · Answer

|dw:1440491573056:dw|

empty · Answer

The picture is nice since all a cube power means is multiply something by itself three times,  $i^3 = i*i*i$ So simplifying this should be no big deal, since you know what $i*i=-1$ already.

jiteshmeghwal9 · Answer

the answer given in book is $42-2i$. nt matching :/

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%282%2B3i%29*%281%2B2i%29-%282%2B3i%29%5E3

ganeshie8 · Answer

must be an algebra error somewhere, just double check..

jiteshmeghwal9 · Answer

yes$$z_1z_2-z_1^3=42-29i-27i^3$$$$=42-29i+27i=42-2i$$

jiteshmeghwal9 · Answer

thanx

ganeshie8 · Answer

In case this is first time, it might be a bit exciting to notice that any power of $i$ always simplifies to one of the numbers :  $\{\pm 1, ~\pm i\}$

ganeshie8 · Answer

This holds :
$$\large  i^{n}\equiv i^{n\pmod{4}}$$

In other words, subtracting/adding $4$ from the exponent doesn't change the number