What is the answer to (z-3/5)^2+(5i)^2=0 ?

The answer to (z-3/5)^2+(5i)^2=0 is z= 28/5 ,z=-22/5

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(z-3/5)^2+(5i)^2=0

(z - \frac{3}{5})^{2} + (5 i)^{2} = 0

z = \frac{28}{5}, z = - \frac{22}{5}

Decimal Notation

z = 5.6, z = - 4.4

Solution steps

(z - \frac{3}{5})^{2} + (5 i)^{2} = 0

Substitute

z = x + y i

(x + y i - \frac{3}{5})^{2} + (5 i)^{2} = 0

Expand

(x + y i - \frac{3}{5})^{2} + (5 i)^{2} : \frac{- 616 + 25 x ^{2} - 25 y ^{2} - 30 x}{25} + i \frac{- 6 y + 10 x y}{5}

\frac{- 616 + 25 x ^{2} - 25 y ^{2} - 30 x}{25} + i \frac{- 6 y + 10 x y}{5} = 0

Rewrite

0

in standard complex form:

0 + 0 i

\frac{- 616 + 25 x ^{2} - 25 y ^{2} - 30 x}{25} + i \frac{- 6 y + 10 x y}{5} = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[\frac{- 616 + 25 x ^{2} - 25 y ^{2} - 30 x}{25} = 0 \frac{- 6 y + 10 x y}{5} = 0]

[\frac{- 616 + 25 x ^{2} - 25 y ^{2} - 30 x}{25} = 0 \frac{- 6 y + 10 x y}{5} = 0] : (x = \frac{28}{5}, x = - \frac{22}{5}, y = 0 y = 0)

Substitute back

z = x + y i

z = \frac{28}{5}, z = - \frac{22}{5}

- i = x^{2} - y^{2} + 2 x y i

i x^{2} - 4 x - 4 i = 0

(x + i y) (3 + 4 i) = 3 - 4 i

x + 1 + i (y + 1) = 2

z^{2} = 2 + i

What is the answer to (z-3/5)^2+(5i)^2=0 ?
The answer to (z-3/5)^2+(5i)^2=0 is z= 28/5 ,z=-22/5