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Question Number 32203 by rahul 19 last updated on 21/Mar/18

Number of solutions of the equation  z^3 +(([3(z^− )^2 ])/(∣z∣))=0 where z is a complex no.

$$\boldsymbol{{N}}{umber}\:{of}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$${z}^{\mathrm{3}} +\frac{\left[\mathrm{3}\left(\overset{−} {{z}}\right)^{\mathrm{2}} \right]}{\mid{z}\mid}=\mathrm{0}\:{where}\:{z}\:{is}\:{a}\:{complex}\:{no}. \\ $$

Commented by rahul 19 last updated on 21/Mar/18

Commented by rahul 19 last updated on 21/Mar/18

pls explain this ′′5 distinct values ofθ′′.  and why r≠−(√3)?

$${pls}\:{explain}\:{this}\:''\mathrm{5}\:{distinct}\:{values}\:{of}\theta''. \\ $$$${and}\:{why}\:{r}\neq−\sqrt{\mathrm{3}}? \\ $$

Commented by MJS last updated on 21/Mar/18

a+bi  r=(√(a^2 +b^2 ))≧0

$${a}+{b}\mathrm{i} \\ $$$${r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqq\mathrm{0} \\ $$

Commented by MJS last updated on 21/Mar/18

  e^(5iθ) =−1  (e^(iθ) )^5 =−1  x^n =c  has got 5 solutions

$$ \\ $$$${e}^{\mathrm{5}{i}\theta} =−\mathrm{1} \\ $$$$\left({e}^{{i}\theta} \right)^{\mathrm{5}} =−\mathrm{1} \\ $$$${x}^{{n}} ={c} \\ $$$$\mathrm{has}\:\mathrm{got}\:\mathrm{5}\:\mathrm{solutions}\: \\ $$

Commented by rahul 19 last updated on 21/Mar/18

thank u sir.

$${thank}\:{u}\:{sir}. \\ $$

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