If-x-p-iq-h-ik-and-p-q-k-h-then-relate-p-q-h-k-R-such-that-x-R-

Question Number 199290 by ajfour last updated on 31/Oct/23

I f x = \sqrt{p + i q} + \sqrt{h + i k}

a n d \frac{p}{q} \neq \frac{k}{h} t h e n r e l a t e p, q, h, k \in R

s u c h t h a t x \in R .

Commented by Frix last updated on 31/Oct/23

\sqrt{x + y i} = r e^{i θ} with r > 0 \land - \frac{π}{2} < θ ⩽ \frac{π}{2}

\sqrt{x + y i} \notin R \Rightarrow - \frac{π}{2} < θ < 0 \lor 0 < θ < \frac{π}{2}

\Rightarrow

\sqrt{p + q i} = a + b i is in the 1^{st} quadrant

\sqrt{h + k i} = c + d i is in the 4^{th} quadrant

(or the other way round)

\Rightarrow for a, b, c, d > 0 we get

\sqrt{p + q i} = a + b i

\sqrt{h + k i} = c - d i

\sqrt{p + q i} + \sqrt{h + k i} = (a + c) + (b - d) i \in R \Rightarrow d = b

\sqrt{p + q i} = a + b i

\sqrt{h + k i} = c - b i

p + q i = (a^{2} - b^{2}) + 2 a b i \Rightarrow q > 0

h + k i = (c^{2} - b^{2}) - 2 b c i \Rightarrow k < 0

For given p \in R, q \in R^{+}

a = \sqrt{\frac{\sqrt{p^{2} + q^{2}} + p}{2}}; b = \sqrt{\frac{\sqrt{p^{2} + q^{2}} - p}{2}}

We can find h \in R, k \in R^{-}

1 . For given h \in R

c = \sqrt{b^{2} + h}; k = - \sqrt{4 (h - p) b^{2} + q^{2}}

2 . For given k \in R^{-}

c = - \frac{a k}{q}; h = \frac{a^{2} k^{2}}{q^{2}} - b^{2}

3 . For given r > b^{2}

c = \sqrt{r - b^{2}}; h = r - 2 b^{2}; k = - 2 b \sqrt{r - b^{2}}

4 . For given - π < θ < 0

c = - b \cot \frac{θ}{2}; h = \frac{2 b^{2} \cos θ}{1 - \cos θ}; k = 2 b^{2} \cot \frac{θ}{2}

I^{'} m not sure what kind of function you

want \dots

Answered by AST last updated on 31/Oct/23

\sqrt{p + i q} = a + i b

\Rightarrow p + i q = a^{2} - b^{2} + 2 i a b

\sqrt{h + i k} = c + d i \Rightarrow \sqrt{p + i q} + \sqrt{h + i k} = (a + c) + (b + d) i

\Rightarrow b = - d

\Rightarrow \sqrt{h + i k} = c - i b \Rightarrow h + i k = c^{2} - b^{2} - 2 i b c

\Rightarrow h = (c^{2} - b^{2}); k = - 2 b c; p = a^{2} - b^{2}; q = 2 a b

Commented by ajfour last updated on 31/Oct/23

t h e r e e x i s t s a z u c h r e l a t i o n

t o u g h t o a r r i v e a t . I w a n t

f (p, q, h, k) = 0 . T h a n k y o u s t i l l .

Answered by mr W last updated on 31/Oct/23

\sqrt{p + q i} = {[\sqrt{p^{2} + q^{2}} (\frac{p}{\sqrt{p^{2} + q^{2}}} + \frac{q}{\sqrt{p^{2} + q^{2}}} i)]}^{\frac{1}{2}}

= {[\sqrt{p^{2} + q^{2}} e^{(\tan^{- 1} \frac{q}{p}) i}]}^{\frac{1}{2}}

= {(p^{2} + q^{2})}^{\frac{1}{4}} e^{\frac{1}{2} (\tan^{- 1} \frac{q}{p}) i}

s i m i l a r l y

\sqrt{h + k i} = {(h^{2} + k^{2})}^{\frac{1}{4}} e^{\frac{1}{2} (\tan^{- 1} \frac{k}{h}) i}

x = \sqrt{p + q i} + \sqrt{h + k i} = {(p^{2} + q^{2})}^{\frac{1}{4}} e^{\frac{1}{2} (\tan^{- 1} \frac{q}{p}) i} + {(h^{2} + k^{2})}^{\frac{1}{4}} e^{\frac{1}{2} (\tan^{- 1} \frac{k}{h}) i}

s u c h t h a t x \in R,

\Rightarrow {(p^{2} + q^{2})}^{\frac{1}{4}} \sin \frac{\tan^{- 1} \frac{q}{p}}{2} + {(h^{2} + k^{2})}^{\frac{1}{4}} \sin \frac{\tan^{- 1} \frac{k}{h}}{2} = 0

Commented by ajfour last updated on 31/Oct/23

g r e a t w a y, t h a n k s i m m e n s e l y s i r!

Answered by MM42 last updated on 31/Oct/23

p = r c o s a & q = r s i n a & r = \sqrt{p^{2} + q^{2}}

h = r^{'} c o s b & k = r^{'} s i n b & r^{'} = \sqrt{h^{2} + k^{2}}

\Rightarrow x = \sqrt{r} e^{i \frac{a}{2}} + \sqrt{r^{'}} e^{i \frac{b}{2}}

\Rightarrow x^{2} = {r e}^{i a} + r^{^{'}} e^{i b} + 2 \sqrt{{r r}^{'}} e^{i (\frac{a + b}{2})}

= r (c o s a + i s i n a) + r^{'} (c o s b + i s i n b) + 2 \sqrt{{r r}^{'}} (c o s (\frac{a + b}{2}) + i s i n (\frac{a + b}{2}))

x \in R \Rightarrow x^{2} \in R \Rightarrow r s i n a + r^{'} s i n b + 2 \sqrt{{r r}^{'}} s i n (\frac{a + b}{2}) = 0

\Rightarrow r s i n a + r^{'} s i n b + 2 \sqrt{{r r}^{'}} \times \sqrt{\frac{1 - c o s (a + b)}{2}} = 0

\Rightarrow p + q + 2 \sqrt{\frac{{r r}^{'} - (r s i n a) (r^{'} c o s b) + (r^{'} s i n b) (r c o s a)}{2}} = 0

\Rightarrow p + q + 2 \sqrt{\frac{{r r}^{'} - q h + p k}{2}} = 0

\Rightarrow p + q + \sqrt{2 (\sqrt{(p^{2} + q^{2}) (h^{2} + k^{2})} + p k - q h)} = 0

Commented by mr W last updated on 31/Oct/23

c a n w e s a y

x \in R \Leftrightarrow x^{2} \in R ?

i t h i n k n o . e x a m p l e : x^{2} = - 2 \in R, b u t

x = \pm \sqrt{2} i \notin R .

Commented by MM42 last updated on 31/Oct/23

S i r W

i f “ x \in R^{″} \Rightarrow x^{2} \in R

b u t “ x^{2} \in R^{″} ⇏ x \in R

i d i d n o t g e t s u c h a r e s u l t

“ x^{2} \in R \Rightarrow x \in R^{″}

Commented by MM42 last updated on 31/Oct/23

y e s r e s u l t o f t h e l a s t l i n e w a s i n c o r r e c t .

t h e r e s u l t i s t h e s a m e l i n e b e f o r e

t h e e n d .

b u t “ i f z^{2} = 0 \Rightarrow z = 0^{″}

Commented by mr W last updated on 31/Oct/23

w i t h

\sqrt{p + q + \sqrt{2 \sqrt{(p^{2} + q^{2}) (h^{2} + k^{2})} + p k - q h}} = 0 ✓

w e e n s u r e t h a t x^{2} \in R . b u t a r e w e a l s o

s u r e t h a t x \in R ?

Answered by Mathspace last updated on 31/Oct/23

p + i q = \sqrt{p^{2} + q^{2}} e^{i a r c t a n (\frac{q}{p})}

a n d \sqrt{p + i q} = {(p^{2} + q^{2})}^{\frac{1}{4}} e^{\frac{1}{2} i a r c t a n (\frac{q}{p})}

a l s o

\sqrt{h + i k} = {(h^{2} + k^{2})}^{\frac{1}{4}} e^{\frac{1}{2} i a r c t a n (\frac{k}{h})}

\Rightarrow x =^{4} \sqrt{p^{2} + q^{2}} {c o s (\frac{1}{2} a r c t a n (\frac{q}{p})

+ i s i n (\frac{1}{2} a r c t a n (\frac{q}{p})}

+^{4} \sqrt{h^{2} + k^{2}} {c o s (\frac{1}{2} a r c t a n (\frac{k}{h}))

+ i s i n (\frac{1}{2} a r c t a n (\frac{k}{h})}

a n d x \in R \Rightarrow I m (\dots .) = 0 \Leftrightarrow

(^{4} \sqrt{p^{2} + q^{2}}) s i n (\frac{1}{2} a r c t a n (\frac{q}{p}))

+^{4} \sqrt{h^{2} + k^{2}} s i n (\frac{1}{2} a r c t a n (\frac{k}{h})) = 0 \Rightarrow

\frac{s i n (\frac{1}{2} a r c t a n (\frac{q}{p})}{s i n (\frac{1}{2} a r c t a n (\frac{k}{h})} = -^{4} \sqrt{\frac{h^{2} + k^{2}}{p^{2} + q^{2}}}

Answered by ajfour last updated on 01/Nov/23

x^{2} = p + h + i (q + k) + 2 \sqrt{w}

\sqrt{w} = \sqrt{(p + i q) (h + i k)} = s - i (q + k)

\Rightarrow p h - q k + i (p k + h q)

= s^{2} - {(q + k)}^{2} - 2 i s (q + k)

p h - q k = s^{2} - {(q + k)}^{2}

2 s = \frac{p k + h q}{k + q}

4 (p h - q k) + 4 {(q + k)}^{2} = {(\frac{p k + h q}{k + q})}^{2}

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