If $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6}$,then $z =$
- A$2\sqrt{3} - 2i$
- B$2\sqrt{3} + 2i$
- C$-2\sqrt{3} + 2i$
- D$-\sqrt{3} + i$
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Similar Questions
Let $z_1$ and $z_2$ be two non-zero complex numbers. Then
If $arg(z - a) = \frac{\pi}{4}$,where $a \in R$,then the locus of $z \in C$ is a
Medium
View SolutionIf the roots of the equation $z^2-i=0$ are $\alpha$ and $\beta$,then $|\operatorname{Arg} \beta-\operatorname{Arg} \alpha|=$
Match the items of List-$I$ with those of List-$II$:
List-$I$ (Complex number) List-$II$ (Polar form) $(i) \sqrt{3}-i$ $(a) 2 \operatorname{cis} \frac{\pi}{6}$ $(ii) \sqrt{3}+i$ $(b) 2 \operatorname{cis} \frac{5 \pi}{6}$ $(iii) -\sqrt{3}+i$ $(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$ $(iv) -\sqrt{3}-i$ $(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$
The correct matching is:
| List-$I$ (Complex number) | List-$II$ (Polar form) |
| $(i) \sqrt{3}-i$ | $(a) 2 \operatorname{cis} \frac{\pi}{6}$ |
| $(ii) \sqrt{3}+i$ | $(b) 2 \operatorname{cis} \frac{5 \pi}{6}$ |
| $(iii) -\sqrt{3}+i$ | $(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$ |
| $(iv) -\sqrt{3}-i$ | $(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$ |
The correct matching is:
Convert the given complex number in polar form: $i$
Medium
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