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Question: The minimum value of \(\frac{3}{2} - 2i\)is....

The minimum value of 32−2i\frac{3}{2} - 2iis.

A

0

B

−2+32i- 2 + \frac{3}{2}i

C

z1 and z2z_{1}\text{ and }z_{2}

D

2/3

Answer

z1 and z2z_{1}\text{ and }z_{2}

Explanation

Solution

Given expression, rsin⁡θ=−3r\sin\theta = - \sqrt{3},minimum value of ∴\thereforeis 0 at tan⁡θ=−35⇒θ=tan⁡−1(−35)\tan\theta = - \frac{\sqrt{3}}{5} \Rightarrow \theta = \tan^{- 1}\left( - \frac{\sqrt{3}}{5} \right). So value of given expression 1+i1−i=1+i1−i×1+i1+i=(1+i)22\frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + i)^{2}}{2} minimum value of 1+i=r(cos⁡θ+isin⁡θ)⇒rcos⁡θ=1,rsin⁡θ=11 + i = r(\cos\theta + i\sin\theta) \Rightarrow r\cos\theta = 1,r\sin\theta = 1 is 0, at r=2,θ=π/4r = \sqrt{2},\theta = \pi/4. So value of given expression ∴\therefore. So minimum value of given expression is 1+i=2(cos⁡π4+isin⁡π4)1 + i = \sqrt{2}\left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right).