## jiteshmeghwal9 one year ago If$z_1=2+3i$$z_2=1+2i$then $z_1z_2-z_1^3=?$

1. Empty

2. jiteshmeghwal9

$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$

3. jiteshmeghwal9

now wht next ?

4. ganeshie8

Nice! $$i^3$$ simplifies further right ?

5. Empty

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

6. jiteshmeghwal9

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

7. ganeshie8

|dw:1440491573056:dw|

8. Empty

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.

9. jiteshmeghwal9

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

10. ganeshie8
11. ganeshie8

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

12. jiteshmeghwal9

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

13. jiteshmeghwal9

thanx

14. ganeshie8

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 : $$\{\pm 1, ~\pm i\}$$

15. ganeshie8

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

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