If z2-2-3i2-i-3-i4- then z is

Z(2 2√(3)i)^2=i(√(3)+i)^4 ⇒(z(2 2√(3)i)^2)=(i(√(3)+i)^4) ⇒(z)+2(2 2√(3)i)=(i)+4(√(3)+i)+2kπ,k∈ℤ ⇒(z)+2( π3)=π2+4(π6)+2kπ,k∈ℤ ⇒(z)=11π6+2kπ,k∈ℤ If nbsp; k= 1 ( ...

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Standard XIII

Mathematics

Modulus of a Complex Number

If z2-2√3i2=i...

Question

If $z (2 - 2 \sqrt{3} i)^{2} = i (\sqrt{3} + i)^{4}$ , then $arg (z)$ is

\frac{5 π}{6}

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\frac{π}{6}

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- \frac{π}{6}

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\frac{2 π}{3}

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Solution

The correct option is C $- \frac{π}{6}$
$z (2 - 2 \sqrt{3} i)^{2} = i (\sqrt{3} + i)^{4}$
$\Rightarrow arg (z (2 - 2 \sqrt{3} i)^{2}) = arg (i (\sqrt{3} + i)^{4}) \Rightarrow arg (z) + 2 arg (2 - 2 \sqrt{3} i) = arg (i) + 4 arg (\sqrt{3} + i) + 2 k π, k \in Z \Rightarrow arg (z) + 2 (- \frac{π}{3}) = \frac{π}{2} + 4 (\frac{π}{6}) + 2 k π, k \in Z \Rightarrow arg (z) = \frac{11 π}{6} + 2 k π, k \in Z$
If $k = - 1$
$arg (z) = - \frac{π}{6} \in (- π, π]$
$∴ arg (z) = - \frac{π}{6}$

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Modulus of a Complex Number

Standard XIII Mathematics

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If z(2−2√3i)2=i(√3+i)4, then arg(z) is

The correct option is C −π6z(2−2√3i)2=i(√3+i)4 ⇒arg(z(2−2√3i)2)=arg(i(√3+i)4)⇒arg(z)+2arg(2−2√3i)=arg(i)+4arg(√3+i)+2kπ,k∈Z⇒arg(z)+2(−π3)=π2+4(π6)+2kπ,k∈Z⇒arg(z)=11π6+2kπ,k∈Z If k=−1 arg(z)=−π6∈(−π,π] ∴arg(z)=−π6

If $z (2 - 2 \sqrt{3} i)^{2} = i (\sqrt{3} + i)^{4}$ , then $arg (z)$ is