(1) Let z = (√(3+4i)) + (√(−3−4i)) , where square root is taken with positive real part. Then Re(z) is _ (a)3 (b) 4 (c) 2 (d) 1 (2)Suppose a,b,c are in AP and a^2 ,b^2 ,c^2 are in GP. If a<b<c and a+b+c=(3/2), then a = _ (a) (1/(2(√2))) (b) (1/(2(√3))) (c) (((√3)−2)/(2(√3))) (d) (((√2)−2)/(2(√2))) (3)A man sent 7 letters to his 7 friend . The letters are kept in addressed envelopes at random. The probability that 3 friend receive correct letters and 4 letters go to wrong destination is _ (a) (1/8) (b) (1/(16)) (c) (3/(32)) (d) (5/(64))


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Question Number 112364 by john santu last updated on 07/Sep/20

$(1) L e t z = \sqrt{3 + 4 i} + \sqrt{- 3 - 4 i}, w h e r e$ $s q u a r e r o o t i s t a k e n w i t h p o s i t i v e$ $r e a l p a r t . T h e n R e (z) i s_$ $(a) 3 (b) 4 (c) 2 (d) 1$ $(2) S u p p o s e a, b, c a r e i n A P a n d a^{2}, b^{2}, c^{2}$ $a r e i n G P . I f a < b < c a n d a + b + c = \frac{3}{2},$ $t h e n a =_$ $(a) \frac{1}{2 \sqrt{2}} (b) \frac{1}{2 \sqrt{3}} (c) \frac{\sqrt{3} - 2}{2 \sqrt{3}} (d) \frac{\sqrt{2} - 2}{2 \sqrt{2}}$ $(3) A m a n s e n t 7 l e t t e r s t o h i s 7 f r i e n d$ $. T h e l e t t e r s a r e k e p t i n a d d r e s s e d$ $e n v e l o p e s a t r a n d o m . T h e p r o b a b i l i t y$ $t h a t 3 f r i e n d r e c e i v e c o r r e c t l e t t e r s$ $a n d 4 l e t t e r s g o t o w r o n g d e s t i n a t i o n$ $i s_$ $(a) \frac{1}{8} (b) \frac{1}{16} (c) \frac{3}{32} (d) \frac{5}{64}$
	Answered by MJS_new last updated on 07/Sep/20

	$\sqrt{z} is unique$ $\sqrt{3 + 4 i} = 2 + i$ $\sqrt{- 3 - 4 i} = 1 - 2 i$ $\Rightarrow z = 3 - i$
	Commented by MJS_new last updated on 07/Sep/20

	$x^{2} = 4$ ${we}^{'} re looking for a l l p o s s i b l e numbers x which$ $solve the equation \Rightarrow x = \pm 2$ $2^{2} we just square without caring if or if not$ ${(any other number)}^{2} = 2^{2} = 4$ $\sqrt{x} = 2$ ${we}^{'} re looking for a l l p o s s i b l e numbers x which$ $solve the equation \Rightarrow x = 4 (try to find another$ $solution x \in C)$ $\sqrt{4} we just extract the root without caring if$ $or if not \sqrt{any other number} = \sqrt{4} = 2$ $so {there}^{'} s no equation given, nothing to$ $solve when we just calculate 7^{2} or \sqrt{49}$ $btw . {there}^{'} s no solution to \sqrt{x} = - 4 (try it)$ $z = r e^{i θ} with r ⩾ 0 and θ \in R$ $z^{q} = r^{q} e^{i q θ} nothing to solve with z, q given$ ${there}^{'} s an exception made to keep geometric$ $problems real :$ $z^{\frac{1}{n}} with z < 0 and n = 2 k + 1 \land k \in Z$ $\Rightarrow z = - {(- z)}^{\frac{1}{n}}$ $or z = r e^{i π} \Rightarrow z^{\frac{1}{2 k + 1}} = r^{\frac{1}{2 k + 1}} e^{i π} n o t r^{\frac{1}{2 k + 1}} e^{\frac{i π}{2 k + 1}}$
	Commented by john santu last updated on 08/Sep/20

	$g r e a t t$
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