jiteshmeghwal9
  • jiteshmeghwal9
If\[z_1=2+3i\]\[z_2=1+2i\]then \[z_1z_2-z_1^3=?\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jiteshmeghwal9
  • jiteshmeghwal9
\[z_1.z_2=(2+3i)(1+2i)\]\[z_1z_2=(2-6)+i(4+3)\]\[z_1z_2=-4+7i\]now\[z_1^3=8+27i^3+54i^2+36i\]\[z_1^3=-46+27i^3+36i\]\[z_1z_2-z_1^3=42-29i+27i^3\]
jiteshmeghwal9
  • jiteshmeghwal9
now wht next ?

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More answers

ganeshie8
  • ganeshie8
Nice! \(i^3\) simplifies further right ?
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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
  • jiteshmeghwal9
\(42-29i-27i\)=\(42-56i\)
ganeshie8
  • ganeshie8
|dw:1440491573056:dw|
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  • 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.
jiteshmeghwal9
  • jiteshmeghwal9
the answer given in book is \(42-2i\). nt matching :/
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=%282%2B3i%29*%281%2B2i%29-%282%2B3i%29%5E3
ganeshie8
  • ganeshie8
must be an algebra error somewhere, just double check..
jiteshmeghwal9
  • jiteshmeghwal9
yes\[z_1z_2-z_1^3=42-29i-27i^3\]\[=42-29i+27i=42-2i\]
jiteshmeghwal9
  • jiteshmeghwal9
thanx
ganeshie8
  • 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 : \(\{\pm 1, ~\pm i\}\)
ganeshie8
  • 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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